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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions - Free PDF Download

In Maths NCERT Solutions Class 10 Chapter 5, students will learn about the arithmetic progression. The NCERT Solutions for Class 10 Maths Chapter 5 PDF file, available for free, can help students to score good marks. Students can download this PDF file by visiting Vedantu. This file is prepared by the best academic experts in India. Every answer is written according to the guidelines set by CBSE. Further, every single step is taken to ensure that students can score good marks.

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Glance of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions | Vedantu

  • An arithmetic progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference d where (d=a2a1).

  • The first term of an AP is denoted by a and the nth term by an.

  • You can find nth term in the AP using the formula: an=a+(n1)d

  • The sum of the first n terms Sn in an AP is calculated using the formula: Sn=n2[2a+(n1)d]

  • The common difference (d) can be positive, negative, or zero.

  • Positive D: The sequence is increasing. (e.g., 2, 5, 8, 11, ...)

  • Negative D: The sequence is decreasing. (e.g., 8, 5, 2, -1, ...)

  • Zero D: The sequence is constant. (e.g., 4, 4, 4, 4, ...)

  • The graph of an AP is a straight line. The slope of this line is equal to the common difference (d).

  • The average of the first and last term is equal to the middle term if there are an even number of terms.

  • This article contains chapter notes, formula and exercises link and important questions for chapter 5 - Arithmetic Progressions.

  • There are four exercises (49 fully solved questions) in class 10th maths chapter 5 Arithmetic Progressions.


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Exercises under NCERT Class 10 Maths Chapter 5 - Arithmetic Progressions

Exercise 5.1 - This exercise contains four problems, each with multiple parts. These problems aim to introduce students to the fundamental formulas of Arithmetic Progressions. These formulas include those for finding the first and last terms, calculating the sum of an A.P., and finding an unknown term of an A.P. using the common difference, among others.


Exercise 5.2 - This is the second exercise contains 20 problems. This exercise covers several problems that involve finding the terms of an A.P. by using formulas and substituting known values of the A.P. The exercise also includes numerous word problems to help students understand how to apply A.P. formulas.


Exercise 5.3 - It contains 20 problems related to arithmetic progressions. The problems in this exercise range from very easy to difficult word problems. To solve these problems, students are advised to have a thorough understanding of all the A.P. formulas provided in the chapter.


Exercise 5.4 - It is the final exercise and contains five problems. The first two problems require students to find the first negative term of a given A.P. or the first term of a given A.P. from the sum and product of two other known terms of the same A.P. The remaining problems are word problems that involve basic applications of volume, length, and other physical measurements.


Access NCERT Solutions for Maths  Chapter 5 – Arithmetic Progression

Exercise 5.1

1. In Which of the Following Situations, Does the List of Numbers Involved Make As Arithmetic Progression and Why?


(i). The Taxi Fare After Each Km When the Fare is Rs 15 for the First Km and Rs 8 for Each Additional Km.

Ans: Given the fare of first km is Rs.15 and the fare for each additional km is Rs. 8. Hence, 

Taxi fare for 1st km is Rs. 15.  

Taxi fare for 2nd km is Rs. 15+8=23.  

Taxi fare for 3rd km is Rs. 23+8=31.  

Similarly, Taxi fare for nth km is Rs. 15+(n1)8

Therefore, we can conclude that the above list forms an A.P with common difference of 8.


(ii). The Amount of Air Present in a Cylinder When a Vacuum Pump Removes a Quarter of the Air Remaining in the Cylinder at a Time.

Ans: Let the initial volume of air in a cylinder be V liter. In each stroke, the vacuum pump removes 14 of air remaining in the cylinder at a time. Hence, 

Volume after 1st stroke is 3V4.   

Volume after 2nd stroke is 34(3V4).   

Volume after 3rd stroke is (34)2(3V4).   

Similarly, Volume after nth stroke is (34)nV.   

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by 34. Therefore, we can conclude that the above list does not forms an A.P.


(iii). The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.

Ans: Given the cost of digging for the first meter is Rs.150 and the cost for each additional meter is Rs. 50. Hence, 

Cost of digging for 1st meter is Rs. 150.  

Cost of digging for 2nd meter is Rs. 150+50=200.  

Cost of digging for 3rd meter is Rs. 200+50=250.  

Similarly, Cost of digging for nth meter is Rs. 150+(n1)50

Therefore, we can conclude that the above list forms an A.P with common difference of 50.


(iv). The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Ans: Given the principal amount is Rs.10000 and the compound interest is 8% per annum. Hence, 

Amount after 1st year is Rs. 10000(1+8100).  

Amount after 2nd year is Rs. 10000(1+8100)2.  

Amount after 3rd year is Rs. 10000(1+8100)3.  

Similarly, Amount after nth year is Rs. 10000(1+8100)n.

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by (1+8100). Therefore, we can conclude that the above list does not forms an A.P.


2. Write first four terms of the A.P. when the first term a and the common difference d are given as follows:


1. a=10,d=10

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(1)

Substituting a=10,d=10 in (1) we get, an=10+10(n1)=10n …..(2)

Therefore, from (2)

a1=10, a2=20, a3=30 and a4=40.


2. a=2,d=0

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(1)

Substituting a=2,d=0 in (1) we get, an=2+0(n1)=2 …..(2)

Therefore, from (2)

a1=2, a2=2, a3=2 and a4=2.


3. a=4,d=3

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(1)

Substituting a=4,d=3 in (1) we get, an=43(n1)=73n …..(2)

Therefore, from (2)

a1=4, a2=1, a3=2 and a4=5.


4. a=1,d=1/2

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(1)

Substituting a=1,d=1/2 in (1) we get, an=1+12(n1)=n32 …..(2)

Therefore, from (2)

a1=1, a2=12, a3=0 and a4=12.


5. a=1.25,d=0.25

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(1)

Substituting a=1.25,d=0.25 in (1) we get, an=1.250.25(n1)=10.25n …..(2)

Therefore, from (2)

a1=1.25, a2=1.5, a3=1.75 and a4=2.


3. For the following A.P.s, write the first term and the common difference.


1. 3,1,1,3,...

Ans: From the given AP, we can see that the first term is 3.

The common difference is the difference between any two consecutive numbers of the A.P.

Common difference = 2nd term1st term

Common difference = 13=2.


2. 5,1,3,7,...

Ans: From the given AP, we can see that the first term is 5.

The common difference is the difference between any two consecutive numbers of the A.P. 

Common difference = 2nd term1st term

Common difference = 1(5)=4.


3. 13,53,93,133,...

Ans: From the given AP, we can see that the first term is 13.

The common difference is the difference between any two consecutive numbers of the A.P. 

Common difference = 2nd term1st term

Common difference = 5313=43.


4. 0.6,1.7,2.8,3.9,...

Ans: From the given AP, we can see that the first term is 0.6.

The common difference is the difference between any two consecutive numbers of the A.P. 

Common difference = 2nd term1st term

Common difference = 1.70.6=1.1.


4. Which of the following are AP’s? If they form an AP, find the common difference d and write three more terms.


1. 2,4,8,16...

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=42=2              …..(1)

a3a2=84=4              …..(2)

a4a3=168=8              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.


2. 2,52,3,72,...

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=522=12              …..(1)

a3a2=352=12              …..(2)

a4a3=723=12              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term 2 and common difference 12.

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(4)

Substituting a=2,d=12 in (1) we get, an=2+12(n1)=n+32 …..(5)

Therefore, from (5)

a5=4, a6=92 and a7=5.


3. 1.2,3.2,5.2,7.2...

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=3.21.2=2              …..(1)

a3a2=5.23.2=2              …..(2)

a4a3=7.25.2=2              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term 1.2 and common difference 2.

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(4)

Substituting a=1.2,d=2 in (1) we get, an=1.2+2(n1)=2n0.8 …..(5)

Therefore, from (5)

a5=9.2, a6=11.2 and a7=13.2.


4. 10,6,2,2,...

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=6(10)=4              …..(1)

a3a2=2(6)=4              …..(2)

a4a3=2(2)=4              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term 10 and common difference 4.

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(4)

Substituting a=10,d=4 in (1) we get, an=10+4(n1)=4n14 …..(5)

Therefore, from (5)

a5=6, a6=10 and a7=14.


5. 3,3+2,3+22,3+32,...

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=(3+2)(3)=2              …..(1)

a3a2=(3+22)(3+2)=2              …..(2)

a4a3=(3+32)(3+22)=2              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term 3 and common difference 2

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(4)

Substituting a=3,d=2 in (1) we get, an=3+(n1)2   …..(5)

Therefore, from (5)

a5=3+42, a6=3+52 and a7=3+62.


6. 0.2,0.22,0.222,0.2222.....

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=0.220.2=0.02              …..(1)

a3a2=0.2220.22=0.002              …..(2)

a4a3=0.22220.222=0.0002              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.


7. 0,4,8,12....

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=40=4              …..(1)

a3a2=8(4)=4              …..(2)

a4a3=12(8)=4              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term 0 and common difference 4

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(4)

Substituting a=0,d=4 in (1) we get, an=04(n1)=44n …..(5)

Therefore, from (5)

a5=16, a6=20 and a7=24.


8. 12,12,12,12....

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=(12)(12)=0              …..(1)

a3a2=(12)(12)=0              …..(2)

a4a3=(12)(12)=0              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term 12 and common difference 0

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(4)

Substituting a=12,d=0 in (1) we get, an=12+0(n1)=12 …..(5)

Therefore, from (5)

a5=12, a6=12 and a7=12.


9. 1,3,9,27,

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=31=2         …..(1)

a3a2=93=6           …..(2)

a4a3=279=18           …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.


10. a,2a,3a,4a,

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=2aa=a              …..(1)

a3a2=3a2a=a              …..(2)

a4a3=4a3a=a              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term a and common difference a

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d       …..(4)

Substituting, a=a,d=ain (4)

we get, an=a+(n1)d....(5)

a5=a+(51)a=5a

a6=a+(61)a=6a

a7=a+(71)a=7a

 

11. a,a2,a3,a4,

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.a2a1=a2a=a(a1)              …..(1)

a3a2=a3a2=a2(a1)         …..(2)

a4a3=a4a3=a3(a1)  ……..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.


12. 2,8,18,32,

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=82=222=2     ……(1)

a3a2=188=3222=2   ………(2)

a4a3=3218=4232=2.......(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term a and common difference is given byan=a+(n1)d      …..(4)

Substituting, a = 2d = 2 in (4)

we get, 

a5=2+(51)2=2+42=52=50

a6=2+(61)2=2+52=62=72

a7=2+(71)2=2+62=72=98

 

13. 3,6,9,12,

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=63=3×23=3(21)     ……(1)

a3a2=96=36=3(32)   ………(2)

a4a3=129=233=3(23)……….(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

 

14. 12,32,52,72,

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=3212=8     ……(1)

a3a2=5232=16  ………(2)

a4a3=7252=24……….(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

 

15.12,52,72,73,...

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

a2a1=5212=24 ……(1)

a3a2=7252=24………(2)

a4a3=7372=24  ……….(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term a and common difference is given byan=a+(n1)d      …..(4)

Substituting a = 12, d=24 in (4)

we get, 

a5=12+(51)24=1+(4)24=97

a6=12+(61)24=1+(5)24=121

a7=12+(71)24=1+(6)24=145


Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P.



a

d 

n 

an 

I

7 

3 

8 

.....

II

18 

.....

10 

0 

III

.....

3 

18 

5 

IV

18.9 

2.5 

.....

3.6 

V

3.5 

0 

105 

.....



(i). Ans: Given, the first Term, a=7  ….. (1)

Given, the common Difference, d=3 …..(2)

Given, the number of Terms, n=8  …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

an=7+(81)3

an=7+21

an=28


(ii). Ans: Given, the first Term, a=18  ….. (1)

Given, the nth term, an=0 …..(2)

Given, the number of Terms, n=10  …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

0=18+(101)d

18=9d

d=2


(iii). Ans: Given, the nth term, an=5 ….. (1)

Given, the common Difference, d=3 …..(2)

Given, the number of Terms, n=18  …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

5=a+(181)(3)

5=a51

an=46


(iv) Ans: Given, the first Term, a=18.9  ….. (1)

Given, the common Difference, d=2.5 …..(2)

Given, the nth term, an=3.6 …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

3.6=18.9+(n1)(2.5)

22.5=(n1)(2.5)

9=(n1)

an=10


(v). Ans: Given, the first Term, a=3.5  ….. (1)

Given, the common Difference, d=0 …..(2)

Given, the number of Terms, n=105  …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

an=3.5+(1051)(0)

an=3.5


2. Choose the correct choice in the following and justify

(i). 30th term of the A.P 10,7,4,..., is

  1. 97

  2. 77

  3. 77

  4. 87

Ans: C. 77 

Given, the first Term, a=10  ….. (1)

Given, the common Difference, d=710=3 …..(2)

Given, the number of Terms, n=30  …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, an=10+(301)(3)

an=1087

an=77


(ii). 11th term of the A.P 3,12,2,..., is

  1. 28

  2. 22 

  3. 38

  4. 4812

Ans: B. 22 

Given, the first Term, a=3  ….. (1)

Given, the common Difference, d=12(3)=52 …..(2)

Given, the number of Terms, n=11  …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, an=3+52(111)

an=3+25

an=22


3. In the following APs find the missing term in the blanks

(i). 2,__,26 

Ans: Given, first term a=2 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     

Substituting the values from (1) we get, an=2+(n1)d  …..(2)

Given, third term a3=26. From (2) we get, 

26=2+(31)d

26=2+2d 

d=12  ….(3)

From (1), (2) and (3) we get for n=2 

a2=2+(21)(12)

a2=14

The sequence is 2,14,26.


(ii). __,13,__,3

Ans: Given, second term a2=13 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d …..(2)

Substituting the values from (1) for n=2 we get, 13=a+d  …..(3)

Given, fourth term a4=3. From (2) we get, 

3=a+3d  …..(4)

Solving (3) and (4) by subtracting (3) from (4) we get,

313=(a+3d)(a+d)

10=2d 

d=5  ….(5)

From (3) and (5) we get 

13=a5

a=18 ……(6)

Substituting the values from (5) and (6) in (2) we get,

an=185(n1)   …..(7)

First term, a=18 and third term a3=8

The sequence is 18,13,8,3.


(iii). 5,__,__,912

Ans: Given, first term a=5 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d …..(2)

Substituting the values from (1) in (2) we get, an=5+(n1)d  …..(3)

Given, fourth term a4=912. From (3) we get, 

912=5+(41)d 

912=5+3d 

d=32  ….(4)

From (3) and (4) we get 

an=5+32(n1) ……(5)

Second term, a2=132 and third term a3=8

The sequence is 5,132,8,912.


(iv). 4,__,__,__,__,6

Ans: Given, first term a=4 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d …..(2)

Substituting the values from (1) in (2) we get, an=4+(n1)d  …..(3)

Given, sixth term a6=6. From (3) we get, 

6=4+(61)d 

6=4+5d 

d=2  ….(4)

From (3) and (4) we get 

an=4+2(n1) ……(5)

Second term a2=2, third term a3=0, fourth term a4=2 and fifth term a5=4.

The sequence is 4,2,0,2,4,6


(v). __,38,__,__,__,22

Ans: Given, second term a2=38 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d …..(2)

Substituting the values from (1) for n=2 we get, 38=a+d  …..(3)

Given, sixth term a6=22. From (2) we get, 

22=a+5d  …..(4)

Solving (3) and (4) by subtracting (3) from (4) we get,

2238=(a+5d)(a+d)

60=4d 

d=15  ….(5)

From (3) and (5) we get 

38=a15

a=53 ……(6)

Substituting the values from (5) and (6) in (2) we get,

an=5315(n1)   …..(7)

First term, a=53, second term a3=23, third term a3=8 and fourth term a4=7

The sequence is 53,38,23,8,7,22.


4. Which term of the A.P. 3,8,13,18,... is 78?

Ans: Given, the first Term, a=3  ….. (1)

Given, the common Difference, d=83=5 …..(2)

Given, the nth term, an=78 …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

78=3+5(n1)

75=5(n1)

15=(n1)

n=16 

Therefore, 16th term of this A.P. is 78.


5. Find the number of terms in each of the following A.P.

(i). 7,13,19,...,205

Ans: Given, the first Term, a=7  ….. (1)

Given, the common Difference, d=137=6 …..(2)

Given, the nth term, an=205 …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

205=7+6(n1)

198=6(n1)

33=(n1)

n=34 

Therefore, given A.P. series has 34 terms.


(ii). 18,1512,13,....,47 

Ans: Given, the first Term, a=18  ….. (1)

Given, the common Difference, d=151218=52 …..(2)

Given, the nth term, an=47 …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

47=1852(n1)

65=52(n1)

26=(n1)

n=27 

Therefore, given A.P. series has 27 terms.


6. Check whether 150 is a term of the A.P. 11,8,5,2,...

Ans: Given, the first Term, a=11  ….. (1)

Given, the common Difference, d=811=3 …..(2)

Given, the nth term, an=150 …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

150=113(n1)

161=3(n1)

1613=(n1)

n=1643 

Since n is nor a natural number. Therefore, 150 is not a term of the given A.P. series.


7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Ans: Given, the 11th Term, a11=38  ….. (1)

Given, the 16th Term, a16=73  …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(3)

Substituting the values from (1) in (3) we get,

38=a+(111)d

38=a+10d      …..(4)

Substituting the values from (2) in (3) we get,

73=a+(161)d

73=a+15d      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

7338=(a+15d)(a+10d)

5=35d

d=7   …..(6)

Substituting value from (6) in (4) we get,

38=a+70

a=32   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

an=32+7(n1)  …..(8)

To find the 31st term substitute n=31 in (8) we get, 

a31=32+7(311)

a31=32+210

a31=178

Therefore, the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73 is 178.


8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Ans: Given, the 3rd Term, a3=12  ….. (1)

Given, the 50th Term, a50=106  …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(3)

Substituting the values from (1) in (3) we get,

12=a+(31)d

12=a+2d      …..(4)

Substituting the values from (2) in (3) we get,

106=a+(501)d

106=a+49d      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

10612=(a+49d)(a+2d)

94=47d

d=2   …..(6)

Substituting value from (6) in (4) we get,

12=a+4

a=8   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

an=8+2(n1)  …..(8)

To find the 29th term substitute n=29 in (8) we get, 

a29=8+2(291)

a29=8+56

a29=64

Therefore, the 29th term of the A.P. is 64.


9. If the 3rd and the 9th terms of an A.P. are 4 and 8 respectively. Which term of this A.P. is zero.

Ans: Given, the 3rd Term, a3=4  ….. (1)

Given, the 9th Term, a9=8  …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     …..(3)

Substituting the values from (1) in (3) we get,

4=a+(31)d

4=a+2d      …..(4)

Substituting the values from (2) in (3) we get,

8=a+(91)d

8=a+8d      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

84=(a+8d)(a+2d)

12=6d

d=2   …..(6)

Substituting value from (6) in (4) we get,

4=a4

a=8   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

an=82(n1)  …..(8)

T find the term which is zero, substitute an=0 in (8)

0=82(n1)

8=2(n1)

4=(n1)

n=5

Therefore, given A.P. series has 5th term as zero.


10. If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Ans: Given that the 17th term of an A.P. exceeds its 10th term by 7 i.e., 

a17=a10+7 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     ….. (2)

For 17th term substitute n=17 in (2) i.e., a17=a+16d  ….. (3)

For 10th term substitute n=10 in (2) i.e., a10=a+9d  ….. (4)

Therefore, from (1), (3) and (4) we get, 

a+16d=a+9d+7

7d=7

d=1 

Therefore, the common difference is 1.


11. Which term of the A.P. 3,15,27,39,... will be 132 more than its 54th term?

Ans: Let nth term of A.P. be 132 more than its 54th term i.e.,

an=a54+132 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     ….. (2)

For 54th term substitute n=54 in (2) i.e., a54=a+53d  ….. (3)

Therefore, from (1), (2) and (3) we get, 

a+(n1)d=a+53d+132

(n1)d53d=132

d=132n54  ….. (4)

Now, given A.P. 3,15,27,39,... 

Common difference d=153=12  ….. (5)

Hence, from (4) and (5) we get 12=132n54

n54=11 

n=65 

Therefore, 65th term of the given A.P. will be 132 more than its 54th term.


12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?

Ans: Let 2 A.P.’s be 

a,a+d,a+2d,a+3d,....   …..(1)

b,b+d,b+2d,b+3d,....    …..(2)

(Since common difference is same)

Given that the difference between their 100th term is 100 i.e., 

a100b100=100 …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     ….. (4)

Therefore, from (3) and (4) we get, 

a+(1001)d(b+(1001)d)=100

ab=100  ….. (5)

Similarly, the difference between their 1000th terms is, 

a1000b1000=[a+(1000a)d][b+(1000a)d]

a1000b1000=ab

a1000b1000=100 

Therefore, the difference between their 1000th terms is 100.


13. How many three-digit numbers are divisible by 7?

Ans: First three-digit number that is divisible by 7 is 105 then the next number will be 105+7=112

Therefore, the series becomes 105,112,119,.... 

This is an A.P. having first term as 105 and common difference as 7.

Now, the largest 3 digit number is 999.

Leu us divide it by 7 , to get the remainder.

999=142×7+5 

Therefore, 9995=994 is the maximum possible three-digit number that is divisible by 7.

Also, this will be the last term of the A.P. series.

Hence the final series is as follows: 105,112,119,...,994 

Let 994 be the nth term of this A.P.

Then, an=105+7(n1) 

994=105+7(n1)

889=7(n1)

127=(n1)

n=128 

Therefore, 128 three-digit numbers are divisible by 7.


14. How many multiples of 4 lie between 10 and 250?

Ans: First number that is divisible by 4 and lie between 10 and 250 is 12. The next number will be 12+4=16

Therefore, the series becomes 12,16,20,.... 

This is an A.P. having first term as 12 and common difference as 4.

Now, the largest number in range is 250.

Leu us divide it by 4 to get the remainder.

250=62×4+2 

Therefore, 2502=248 is the last term of the A.P. series.

Hence the final series is as follows: 12,16,20,....,248 

Let 248 be the nth term of this A.P.

Then, an=12+4(n1) 

248=12+4(n1)

236=4(n1)

59=(n1)

n=60 

Therefore, 60 multiples of 4 lie between 10 and 250.


15. For what value of n, are the nth terms of two APs 63,65,67,.... and 3,10,17,.... equal

Ans: Given 2 A.P.’s are

63,65,67,....   …..(1)

Its first term is 63 and common difference is 6563=2 

3,10,17,....    …..(2)

Its first term is 3 and common difference is 103=7 

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     ….. (3)

Therefore, from (1) and (3) we get the nth term of the first A.P. is

an=63+2(n1)

an=61+2n  ….. (4)

And from (2) and (3) we get the nth term of the second A.P. is

bn=3+7(n1)

bn=4+7n  ….. (5)

If the nth terms of two APs 63,65,67,.... and 3,10,17,.... are equal the from (4) and (5),

an=bn

61+2n=4+7n

65=5n

n=13 

Therefore, the 13th term of both the A.P.’s are equal.


16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

Ans: Given the 7th term of A.P. is 12 more than its 5th term i.e.,

a7=a5+12 …..(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     ….. (2)

For 5th term substitute n=5 in (2) i.e., a5=a+4d  ….. (3)

For 7th term substitute n=7 in (2) i.e., a7=a+6d  ….. (4)

Therefore, from (1), (3) and (4) we get, 

a+6d=a+4d+12

2d=12

d=6   ….. (5)

Substituting (5) in (2) we get, an=a+6(n1) ……(6)

Given the third term of the A.P. is 16. Hence from (6),

16=a+6(31)

16=a+12

a=4   ….. (7)

Hence from (6), an=4+6(n1)

Therefore, the A.P. will be 4,10,16,22,.....


17. Find the 20th term from the last term of the A.P. 3,8,13,...,253 

Ans: Given A.P. 3,8,13,...,253. To find the 20th term from the last write the given A.P. in reverse order and then find its 20th term.

Required A.P. is 253,...,13,8,3  ….. (1) 

Its first A.P. is 253 and common difference is 813=5.  …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     ….. (3)

Hence from (2) and (3) we get, an=2535(n1)   …..(4)

Substitute n=20 in (4) we get, 

a20=2535(201)

a20=158.

Therefore, 20th term from the last is 158.


18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

Ans: Given the sum of 4th and 8th terms of an A.P. is 24 i.e.,

a4+a8=24 …..(1)

Given the sum of 6th and 10th terms is 44 i.e.,

a6+a10=44 …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d     ….. (3)

For 4th term substitute n=4 in (3) i.e., a4=a+3d  ….. (4)

For 6th term substitute n=6 in (3) i.e., a6=a+5d  ….. (5)

For 8th term substitute n=8 in (3) i.e., a8=a+7d  ….. (6)

For 10th term substitute n=10 in (3) i.e., a10=a+9d  ….. (7)

Therefore, from (1), (6) and (4) we get, 

(a+3d)+(a+7d)=24

2a+10d=24

a+5d=12   ….. (8)

From (2), (5) and (7) we get, 

(a+5d)+(a+9d)=44

2a+14d=44

a+7d=22   ….. (9)

Subtracting (8) from (9) we get,  

(a+7d)(a+5d)=2212

2d=10

d=5 ….. (10)

Substituting this value from (10) in (9) we get,

a+35=22

a=13 ….. (11)

Therefore from (10) and (11), the first three terms of the A.P. are 13,8,3.


19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs  200 each year. In which year did his income reach Rs 7000?

Ans: Given in the first year, annual salary is Rs 5000.  

In the second year, annual salary is Rs 5000+200=5200.  

In the third year, annual salary is Rs 5200+200=5400

This series will form an A.P. with first term 5000 and common difference 200.  

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d  

Therefore, In the nth year, annual salary is an=5000+200(n1)

an=4800+200n …. (1)

To find the year in which his annual income reaches Rs 7000, substitute an=7000 in (1) and find the value of ni.e., 

7000=4800+200n

2200+200n

n=11 

Therefore, in 11th year i.e., in 2005 his salary will be Rs 7000.


20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

Ans: Given in the first week the savings is Rs 5.  

In the second week the savings is Rs 5+1.75=6.75.  

In the third week the savings is Rs 6.75+1.75=8.5

This series will form an A.P. with first term 5 and common difference 1.75.  

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d  

Therefore, In the nth week the savings is an=5+1.75(n1)

an=3.25+1.75n …. (1)

To find the week in which her savings reaches Rs 20.75, substitute an=20.75 in (1) and find the value of ni.e., 

20.75=3.25+1.75n

17.5=1.75n

n=10 

Therefore, in 10th week her savings will be Rs 20.75.


Exercise 5.3

1. Find the sum of the following APs.

(i). 2,7,12,.... to 10 terms.

Ans: Given, the first Term, a=2  ….. (1)

Given, the common Difference, d=72=5 …..(2)

Given, the number of Terms, n=10  …..(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, Sn=102[2(2)+(101)(5)]

Sn=5[4+45]

Sn=245


(ii). 37,33,29,... to 12 terms

Ans: Given, the first Term, a=37  ….. (1)

Given, the common Difference, d=33(37)=4 …..(2)

Given, the number of Terms, n=12  …..(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, Sn=122[2(37)+(121)(4)]

Sn=6[74+44]

Sn=180


(iii).0.6,1.7,2.8,..... to 100 terms

Ans: Given, the first Term, a=0.6  ….. (1)

Given, the common Difference, d=1.70.6=1.1 …..(2)

Given, the number of Terms, n=100  …..(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, Sn=1002[2(0.6)+(1001)(1.1)]

Sn=50[1.2+108.9]

Sn=5505


(iv). 115,112,110,..... to 11 terms

Ans: Given, the first Term, a=115  ….. (1)

Given, the common Difference, d=112115=160 …..(2)

Given, the number of Terms, n=11  …..(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, Sn=112[2(115)+(111)(160)]

Sn=112[4+530]

Sn=3320


2. Find the sums given below

(i). 7+1012+14+.....+84

Ans: Given, the first Term, a=7  ….. (1)

Given, the common Difference, d=10127=72 …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

an=7+72(n1)=72(n+1)  ….. (4)

Given, last term of the series, an=84  …..(5)

Substituting (5) in (4) we get, 84=72(n+1)

24=(n+1) 

n=23  ……(6)

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, Sn=232[7+84]

Sn=232(91)

Sn=104612


(ii). 34+32+30+.....+10 

Ans: Given, the first Term, a=34  ….. (1)

Given, the common Difference, d=3234=2 …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

an=342(n1)=362n  ….. (4)

Given, last term of the series, an=10  …..(5)

Substituting (5) in (4) we get, 10=362n

2n=26 

n=13  ……(6)

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, Sn=132[34+10]

Sn=132(44)

Sn=286


(iii). 5+(8)+(11)+.....+(230) 

Ans: Given, the first Term, a=5  ….. (1)

Given, the common Difference, d=8(5)=3 …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

an=53(n1)=23n  ….. (4)

Given, last term of the series, an=230  …..(5)

Substituting (5) in (4) we get, 230=23n

228=3n 

n=76  ……(6)

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, Sn=762[5+(230)]

Sn=762(235)

Sn=8930


3. In an AP

(i). Given a=5, d=3, an=50, find n and Sn.

Ans: Given, the first Term, a=5  ….. (1)

Given, the common Difference, d=3 …..(2)

Given, nth term of the A.P., an=50  …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (4)

Substituting the values from (1), (2) and (3) in (4) we get, 

50=5+3(n1)=2+3n 

Simplifying it further we get, 

n=5023 

n=16   …..(5)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(6)

Substituting the values from (1), (2) and (5) in (6) we get, Sn=162[2(5)+(161)(3)]

Sn=8[10+45]

Sn=440


(ii). Given a=7, a13=35, find d and S13.

Ans: Given, the first Term, a=7  ….. (1)

Given, 13th term of the A.P., a13=35  …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

35=7+(131)d=7+12d 

Simplifying it further we get, 

d=2812 

d=73   …..(4)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(5)

Substituting the values from (1) and (4) in (5) we get, S13=132[2(7)+(131)(73)]

S13=132[14+28]

S13=273


(iii). Given d=3, a12=37, find a and S12.

Ans: Given, the common difference, d=3  ….. (1)

Given, 12th term of the A.P., a12=37  …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

37=a+3(121)=a+33 

Simplifying it further we get,  

a=4   …..(4)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(5)

Substituting the values from (1) and (4) in (5) we get, S12=122[2(4)+(121)(3)]

S12=6[8+33]

S12=246


(iv). Given a3=15, S10=125 find a10 and d.

Ans: Given, 3rd term of the A.P., a3=15  …..(1)

Given, the sum of terms, S10=125  ….. (2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1) in (3) we get, 

15=a+(31)d=a+2d  …..(4)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(5)

Substituting the values from (1) in (5) we get, 125=102[2a+(101)d]

125=5[2a+9d]

25=2a+9d  …..(5)

Let us solve equations (4) and (5) by subtracting twice of (4) from (5) we get,

2530=(2a+9d)(2a+4d)

5=5d 

d=1    …..(6)

From (4) and (6) we get, a=17   …..(7)

From (3), (6) and (7) for n=10 we get,

a10=17(101)

a10=8


(v). Given S9=75, d=5 find a and a9.

Ans: Given, common difference, d=5  …..(1)

Given, the sum of terms, S9=75  ….. (2)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(3)

Substituting the values from (1), (2) in (3) we get, 75=92[2a+5(91)]

25=3[a+20]

3a=35

a=353  …..(4)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (5)

Substituting the values from (1), (4) in (5) we get, 

a9=353+5(91) 

a9=353+40 

a9=853


(vi) Given a=2, d=8, Sn=90, find n and an.

Ans: Given, common difference, d=8  …..(1)

Given, first term, a=2  …..(2)

Given, the sum of terms, Sn=90  ….. (3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(3)

Substituting the values from (1), (2), (3) in (4) we get, 90=n2[2(2)+8(n1)]

45=n[2n1]

2n2n45=0

2n210n+9n45=0

2n(n5)+9(n5)=0

(n5)(2n+9)=0

n=5  …..(4)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (5)

Substituting the values from (1), (2), (4) in (5) we get, 

a5=2+8(51) 

a5=2+32 

a5=34


(vii). Given a=8, Sn=210, an=62, find n and d.

Ans: Given, first term, a=8  …..(1)

Given, the sum of terms, Sn=210  ….. (2)

Given, the nth term, an=62 …..(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(4)

Substituting the values from (1), (2) in (4) we get, 210=n2[2(8)+d(n1)]

420=n[16+(n1)d]  …..(4)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (5)

Substituting the values from (1), (3) in (5) we get, 

62=8+(n1)d   …..(6) 

Let us solve equations (4) and (6) by subtracting n times of (6) from (4) we get,

42062n=(16n+n(n1)d)(8n+n(n1)d)

42062n=8n 

420=70n

n=6  ……(7)

Substituting the values from (7) in (6) we get, 

62=8+(61)d

54=5d 

d=545


(viii). Given Sn=14, d=2, an=4, find n and a.

Ans: Given, common difference, d=2  …..(1)

Given, the sum of terms, Sn=14  ….. (2)

Given, the nth term, an=4 …..(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(4)

Substituting the values from (1), (2) in (4) we get, 14=n2[2a+2(n1)]

14=n[a+n1]  …..(5)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (6)

Substituting the values from (1), (3) in (6) we get, 

4=a+2(n1)   …..(7) 

Let us solve equations (5) and (7) by substituting the value of a from (7) in (5) we get,

14=n[(42(n1))+n1]

14=n[5n]

n25n14=0 

n27n+2n14=0

(n7)(n+2)=0

n=7 (Since n cannot be negative)  ……(8)

Substituting the values from (8) in (7) we get, 

4=a+2(71)

4=a+12 

a=8


(ix). Given a=3, n=8, S=192, find d.

Ans: Given, first term, a=3  …..(1)

Given, the sum of terms, Sn=192  ….. (2)

Given, the number of terms, n=8 …..(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(4)

Substituting the values from (1), (2) in (4) we get, 192=82[2(3)+d(81)]

192=4[6+7d]

48=6+7d

42=7d

d=6


(x). Given l=28, S=144 and there are total 9 terms. Find a.

Ans: Given, last term, l=28  …..(1)

Given, the sum of terms, Sn=144  ….. (2)

Given, the number of terms, n=9 …..(3)

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]     …..(4)

Substituting the values from (1), (2) in (4) we get, 144=92[a+28]

32=a+28

a=4


4. How many terms of the A.P. 9,17,25... must be taken to give a sum of 636?

Ans: Given, common difference, d=179=8  …..(1)

Given, first term, a=9  …..(2)

Given, the sum of terms, Sn=636  ….. (3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(3)

Substituting the values from (1), (2), (3) in (4) we get, 636=n2[2(9)+8(n1)]

636=n(5+4n)

4n2+5n636=0

4n2+53n48n636=0

n(4n+53)12(4n+53)=0

(n12)(4n+53)=0

n=12 or 534

Since n can only be a natural number n=12


5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Ans: Given, first term, a=5  …..(1)

Given, the sum of terms, Sn=400  ….. (2)

Given, the nth term, an=45 …..(3)

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]     …..(4)

Substituting the values from (1), (2), (3) in (4) we get, 400=n2[5+45]

400=25n 

n=16 

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (5)

Substituting the values from (1), (3) in (5) we get, 

45=5+(161)d   

40=15d 

d=83


6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Ans: Given, first term, a=17  …..(1)

Given, the common difference, d=9  ….. (2)

Given, the nth term, an=350 …..(3)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (4)

Substituting the values from (1), (2), (3) in (4) we get, 

350=17+9(n1)   

333=9(n1) 

37=(n1)

n=38 ……(5)

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]     …..(6)

Substituting the values from (1), (5), (3) in (6) we get, S38=382[17+350]

S38=19(367) 

S38=6973


7. Find the sum of first 22 terms of an AP in which d=7 and 22nd term is 149.

Ans: Given, the common difference, d=7  ….. (1)

Given, the 22nd term, a22=149 …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

149=a+7(221)   

149=a+147 

a=2  ……(4)

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]     …..(5)

Substituting the values from (1), (2), (4) in (5) we get, S22=222[2+149]

S22=11(151) 

S22=1661


8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Ans: Given, the 2nd term, a2=14  ….. (1)

Given, the 3rd term, a3=18 …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d   ….. (3)

Substituting the values from (1) in (3) we get, 

14=a+d   …..(4)

Substituting the values from (2) in (3) we get, 

18=a+2d   …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

1814=(a+2d)(a+d) 

d=4  ……(6)

Substituting the value from (6) in (4) we get a=10.   …..(7) 

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     …..(8)

Substituting the values from (7), (6) in (8) we get for n=51 ,

 S51=512[2(10)+4(511)]

S51=512[20+200] 

S51=5610


9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Ans: Given, the sum of first 7 terms, S7=49  ….. (1)

Given, the sum of first 17 terms, S17=289   …..(2)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     ….. (3)

Substituting the values from (1) in (3) we get, 

49=72[2a+(71)d]

7=a+3d   …..(4)

Substituting the values from (2) in (3) we get, 

289=172[2a+(171)d]

17=a+8d   …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

177=(a+8d)(a+3d) 

10=5d 

d=2  ……(6)

Substituting the value from (6) in (4) we get a=1.   …..(7) 

Substituting the values from (7), (6) in (3) we get,

 Sn=n2[2+2(n1)]

Sn=n2


10. Show that a1,a2...,an,... form an AP where an is defined as below. Also find the sum of the first 15 terms in each case. 

(i). an=3+4n 

Ans:  Consider two consecutive terms of the given sequence. Say an,an+1. Difference between these terms will be 

an+1an=[3+4(n+1)][3+4n] 

an+1an=4(n+1)4n 

an+1an=4

Which is a constant nN

For n=1, a1=3+4=7 

Therefore, it is an A.P. with first term 7 and common difference 4

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     

Therefore, S15=152[2(7)+4(151)]

S15=152[14(5)] 

S15=525


(ii). an=95n 

Ans:  Consider two consecutive terms of the given sequence. Say an,an+1. Difference between these terms will be 

an+1an=[95(n+1)][95n] 

an+1an=5(n+1)+5n 

an+1an=5

Which is a constant nN

For n=1, a1=95=4 

Therefore, it is an A.P. with first term 4 and common difference 5

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     

Therefore, S15=152[2(4)5(151)]

S15=15[31] 

S15=465


11. If the sum of the first n terms of an AP is 4nn2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms. 

Ans:  Given, the sum of the first n terms of an A.P. is 4nn2.

First term =S1=41=3. …..(1)

Sum of first two terms =S2=8(2)2=4  …..(2) 

From (1) and (2), 2nd term =S2S1=43=1

Sum of first three terms =S3=12(3)2=3  …..(3) 

From (3) and (2), 3rd term =S3S2=34=1

Similarly, 

Sum of first n terms =Sn=4nn2  …..(4) 

Sum of first n1 terms =Sn1=4(n1)(n1)2=n2+6n5  …..(5) 

From (4) and (5), nth term =SnSn1=(4nn2)(n2+6n5)=52n  …..(6)

From (6), 10th term is 52(10)=15.


12. Find the sum of first 40 positive integers divisible by 6

Ans:  First positive integer that is divisible by 6 is 6 itself. 

Second positive integer that is divisible by 6 is 6+6=12.

Third positive integer that is divisible by 6 is 12+6=18.  

Hence, it is an A.P. with first term and common difference both as 6.  

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     

Therefore, for n=40,

S40=402[2(6)+6(401)]

S40=120[41] 

S40=4920


13. Find the sum of first 15 multiples of 8.

Ans:  First positive integer that is divisible by 8 is 8 itself. 

Second positive integer that is divisible by 8 is 8+8=16.

Third positive integer that is divisible by 8 is 16+8=24.  

Hence, it is an A.P. with first term and common difference both as 8.  

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]     

Therefore, for n=15,

S15=152[2(8)+8(151)]

S15=60[16]

S15=960


14. Find the sum of the odd numbers between 0 and 50.

Ans: The odd numbers between 0 and 50 are 1,3,5,...,49

It is an A.P. with first term 1 and common difference 2.  ….(1)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d  … (2)

Substitute an=49 and values from (1) into (2)

49=1+2(n1)

24=(n1) 

n=25  ……(3)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]  …..(4)

Substituting values from (1), (3) in (4) we get,

S25=252[2+2(251)]

S25=25[25] 

S25=625


15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Ans: Penalty of delay for first day is Rs. 200.

Penalty of delay for second day is Rs. 250.

Penalty of delay for third day is Rs. 300.

Hence it is an A.P. with first term 200 and common difference 50.

Money the contractor has to pay as penalty, if he has delayed the work by 30 days is the sum of first 30 terms of the A.P.

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore,

S30=302[2(200)+50(301)]

S30=15[400+50(29)]

S30=27750

Therefore, the contractor has to pay Rs 27750 as penalty.


16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Ans: Let the first prize be of Rs. a then the second prize will be of Rs. a20, the third prize will be of Rs. a40.

Therefore, it is an A.P. with first term a and common difference 20.

Given, S7=700

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore,

S7=72[2a20(71)]

700=7[a60]

100=a60

a=160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120,Rs 100, Rs 80, Rs 60, and Rs 40.


17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Ans: Each section of class I will plant 1 tree each. Therefore, total trees planted by class I are 3.

Each section of class II will plant 2 trees each. Therefore, total trees planted by class II are 3×2=6.

Each section of class III will plant 3 trees each. Therefore, total trees planted by class III are 3×3=9.

Therefore, it is an A.P series with first term and common difference both as 3.

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore,

S12=122[2(3)3(121)]

S12=6[39]

S12=234

Therefore, 234 trees will be planted by the students.


18. A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ......... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?


a spiral made up of thirteen consecutive semicircles


Ans: Length of first semi-circle I1=π(0.5) cm. 

Length of second semi-circle I2=π(1) cm.

Length of third semi-circle I3=π(1.5) cm.

Therefore, it is an A.P series with first term and common difference both as π(0.5)

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore,

S13=132[2(0.5π)+(0.5π)(131)]

S13=7×13×(0.5π)

S13=7×13×12×227 

S13=143

Therefore, the length of such spiral of thirteen consecutive semi-circles

will be 143 cm.


19. The 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?


how many logs are in the top row


Ans: Total logs in first row are 20.  

Total logs in second row are 19.

Total logs in third row are 18.

Therefore, it is an A.P series with first term 20 and common difference 1

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore,

200=n2[2(20)(n1)]

400=n[41n]

n241n+400=0

n216n25n+400=0

n(n16)25(n16)=0

(n16)(n25)=0 

For n=25, after 20th term, all terms are negative, which is illogical as terms are representing the number of logs and number of logs being negative is illogical.

n=16 

Total logs in 16th row =20(161)=5 

Therefore, 200 logs will be placed in 16 rows and the total logs in 16th row will be 5.


20. In a potato race, a bucket is placed at the starting point, which is 5m from the first potato and other potatoes are placed 3m apart in a straight line. There are ten potatoes in the line. 

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

(Hint: to pick up the first potato and the second potato, the total

distance (in metres) run by a competitor is 2×5+2×(5+3))


the total distance the competitor has to run


Ans: Total distance run by competitor to collect and drop first potato =2×5=10m.

Total distance run by competitor to collect and drop second potato =2×(5+3)=16m. 

Total distance run by competitor to collect and drop third potato =2×(5+3+3)=22m. 

Therefore, it is an A.P series with first term 10 and common difference 6

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore, to collect and drop 10 potatoes total distance covered is

S10=102[2(10)+6(101)]

S10=5[74]

S13=370

Therefore, the competitor will run a total distance of 370m.


Exercise 5.4

1. Which term of the A.P. 121,117,113,... is its first negative term?

(Hint: Find n for an<0)

Ans: Given A.P. 121,117,113,... 

Its first term is 121 and common difference is 117121=4

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d.

Therefore the nth term of the given A.P. is an=1214(n1)   ….. (1)

To find negative term, find n such that an<0  

Hence from (1),

1214(n1)<0

121<4(n1) 

1214+1<n

n>1254 

n>31.25 

Therefore, the 32nd term of the given A.P. will be its first negative term.


2. The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Ans: Given the sum of third and seventh term of A.P., a3+a7=6  …..(1)

Given the sum of third and seventh term of A.P., a3a7=8  …..(2)

We know that the nth term of the A.P. with first term a and common difference d is given by an=a+(n1)d. Therefore, 

For n=3,a3=a+2d 

For n=7,a7=a+6d 

From (1), a3+a7=(a+2d)+(a+6d) 

2a+8d=6 

a+4d=3   ….. (3)

From (2), a3a7=(a+2d)(a+6d) 

a2+8ad+12d2=8   …..(4)

Let us now solve equations (3) and (4) by substituting the value of a from (3) into (4).

(34d)2+8d(34d)+12d2=8

924d+16d2+24d32d2+12d2=8 

4d2+1=0

d2=14

d=12,12 …..(5)

CASE 1: For d=12 

Substitute d=12 in (6) we get, a=1 …..(6)

Therefore, it is an A.P series with first term 1 and common difference 12

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore, 

S16=162[2+12(161)]

S16=4[19] 

S16=76 

CASE 2: For d=12 

Substitute d=12 in (6) we get, a=5 …..(7)

Therefore, it is an A.P series with first term 5 and common difference 12 and hence, 

Sn=n2[2a+(n1)d]

S16=162[2(5)12(161)]

S16=4[5] 

S16=20


3. A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 212 m apart, what is the length of the wood required for the rungs? 

(Hint: number of rungs =25025 )


the length of the wood required for the rungs


Ans: Distance between first and last rungs is 212m=52m=250cm.

Distance between two consecutive rungs is 25cm.

Therefore, total number of rungs are 25025+1=11.

Also, we can observe that the length of each rung is decreasing in a uniform order. So, we can conclude that the length of rungs is in A.P. with first term 45, common difference 25 and number of terms 11.

We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]. Therefore, 

S11=112[45+25]

S11=11[35] 

S11=385

Therefore, the length of the wood required for the rungs is 385cm.


4. The houses of a row are number consecutively from 1 to 49. Show that there is a value of x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it.

Find this value of x.

(Hint:Sx1=S49Sx)

Ans: Given houses are numbered 1,2,3,4,....

Clearly, they are numbered in A.P. series with both first term and common difference as 1.

Now, there is house numbered x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it i.e., Sx1=S49Sx


the sum of the number of houses


We know that the sum of n terms of the A.P. with first term a and last term l is given by Sn=n2[a+l]. Therefore, 

Sx1=S49Sx

{(x1)2[1+(x1)]}={492[1+49]}{x2[1+x]}

x(x1)2=49[25]x(x+1)2 

x(x1)=2450x(x+1)

2x2=2450

x=35   (Since house number cannot be negative)

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.


5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 14m and a tread of 12m (See figure) calculate the total volume of concrete required to build the terrace.


the total volume of concrete required to build the terrace


Ans: Given that a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 14m and a tread of 12m. An easy illustration of the problem is depicted below.


An easy illustration of the problem is depicted


Here blue step is the lowermost step. Let it be known as step 1. The volume of step 1 is 12×14×50 m3.

The red step is the second lowermost step. Let it be known as step 2. The volume of step 2 is 12×12×50 m3.

The green step is the third lower step. Let it be known as step 3. The volume of step 3 is 12×1×50 m3.


the height is increasing with each increasing


We can see that the height is increasing with each increasing step by a factor of 14, length and width being constant. Hence the volume of each step is increasing by 12×14×50 m3.

Therefore, we can conclude that the volume of steps is in A.P. with first term and common difference both as 12×14×50=254 m3.

We know that the sum of n terms of the A.P. with first term a and common difference d is given by Sn=n2[2a+(n1)d]. Therefore, 

S15=152[2(254)+(254)(151)]

S15=152(254)[16] 

S15=15252

S15=750

Therefore, volume of concrete required to build the terrace is 750 m3.


NCERT Solutions for Class 10 Maths Chapter 5 solutions PDF download

5.1 Introduction: In this chapter, we'll understand the unique number patterns and will learn how to identify them, finding specific numbers in the given sequence and also how to find the sum of all numbers in a given sequence of numbers.


5.2 Arithmetic Progression: In simple terms, Arithmetic Progression or AP can be defined as a sequence of numbers. This sequence exists in an order in which the difference between any two consecutive numbers would be constant.


For Example: if 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 is a series of natural numbers. It can be said that this series is also an arithmetic progression between the difference between every two successive terms is 1. If we have a similar series of odd and even numbers, still, the difference between two successive terms will be two. This means that the odd and even number series will also be arithmetic progressions.


There are three different types of progressions. These different types of progressions that are mentioned in Chapter 5 Class 10 Maths NCERT book are:


  • Arithmetic Progression (AP)

  • Geometric Progression (GP)

  • Harmonic Progression (HP)


There are also three other definitions of arithmetic progression that students should remember for writing Arithmetic Progression Class 10 NCERT Solutions. These definitions are:


Definition One: Arithmetic progression is a mathematical sequence in which the difference between any two consecutive terms is always a constant. It can also be abbreviated as AP


Definition Two: It is also mentioned in NCERT Solutions Class 10 Maths Chapter 5 that arithmetic progression or sequence is the sequence of numbers. In this sequence of consecutive numbers, it is possible to find the next number by adding a fixed number to the previous number in the chain


Definition Three: In Arithmetic Progression Class 10 Solutions, the common difference of the AP is the fixed number that one should add to any term of the arithmetic progression. For example, in the arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, the value of the common difference is 3


In Class 10 Maths Chapter 5 Solutions, there are three main terms. These terms are:


  • The common difference (d)

  • nth Term (an)

  • The num of the first n terms (Sn)

  • These three terms are used to represent the property of arithmetic progression. In the next section, we will look at these three properties in more detail.


The Common Difference in Arithmetic Progression

For any given series of arithmetic progression, the terms that are used are the first term, the common difference between any two terms, and the nth term. Let’s assume that a1, a2, a3, a4, …, an is an arithmetic progression.


This means that the value of the common difference ‘d’ is:


D=a2a1=a3a2==a1an1


Here, d is the value of the common difference. The value of d can be positive, negative, or zero.


The First Term of Arithmetic Progression


If an individual wants to write the arithmetic progression in terms of its common difference for solving, then it can be written as:


A, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …, a + (n - 1) d


In this sequence, a is the first term of the progression.


The General Form of an Arithmetic Progression

In this section, students will be able to do just that. Before we proceed, a student should begin with an assumption that the arithmetic progression for class 10 maths chapter 5 solutions is a1,a2,a3,,an


Position of Terms

Representation of Terms

Values of Terms

1

a1

A = a + (1 - 1) d

2

a2

A + d = a + (2 - 1) d

3

a3

A + 2d = a + (3 - 1) d

4

a4

A + 3d = a + (4 - 1) d

.

.

.

.

.

.

n

an

A + (n - 1) d



5.3 The nth Term of Arithmetic Progression (AP)

This formula can be used for finding the class 10 maths chapter 5 NCERT solutions in which one needs to get the value of the nth term of an arithmetic progression. The formula can be written as:


An=a+(n1)d


Here, a is the first term, d is the value of the common difference, n is the number of terms, and an is the nth term.


Let’s take an example. Try to find out the nth term of the following arithmetic progression 1, 2, 3, 4, 5, …, an. The total number of terms is 15.


We know that n = 15. This means that according to the formula, we can say that:


An=a+(n1)d


Since, a = 1, and the common difference or d = 2 - 1 = 1


Then, an=1+(151)1=1+14=15.


The behaviour of the entire sequence will also depend on the values of the common difference. This means that if the value of the common difference is positive, then the member terms will grow towards positive infinity. And if the value of the common difference is negative, then the member terms will move towards negative infinity.


5.4 The Sum of the First n Terms of an Arithmetic Progression(AP)

One can easily calculate the sum of n terms of any known progression. For an arithmetic progression, it is possible to calculate the sum of the first n terms if the value of the first term and the total terms are known. The formula is mentioned below.


S=n2


2a+(n1)×d


But what if the value of the last term of the arithmetic progression is given? In that case, students should use the formula that is mentioned below.


S = n2(first term + last term)


For ease of revision, we have also summarized all the major formulas of this chapter in a table. That table is mentioned below.


General Form of AP

A, a + d, a + 2d, a + 3d, a + 4d, …, a + nd

The nth term of AP

An=a+(n1)×d

Sum of n terms in AP

S=n2(2a+(n1)×d)

Sum of all terms in a finite AP with the last term as I

N2(a+I)



Class 10 Maths Chapter 5: Exercises Breakdown

Chapter 5 -Arithmetic Progressions All Exercises in PDF Format

Exercise 5.1

4 Questions & Solutions

Exercise 5.2

20 Questions & Solutions

Exercise 5.3

20 Questions & Solutions

Exercise 5.4

5 Questions & Solutions



Conclusion

NCERT Solutions for Class 10 Maths Chapter on Arithmetic Progressions provided by Vedantu offer a comprehensive understanding of this fundamental concept.Students can gain a solid understanding of arithmetic progressions by concentrating on important concepts such as calculating the sum of terms, nth term, and common difference. It's important to pay close attention to the step-by-step solutions provided in the NCERT Solutions, as they help clarify concepts and reinforce problem-solving techniques.Arithmetic progressions are fundamental to more advanced mathematical ideas, hence it's important to understand them. This chapter has usually been the subject of five or six questions on previous year's exam papers. Therefore, practicing a variety of problems from NCERT Solutions and previous year papers can enhance preparation and confidence for exams.




Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

1. I’m having doubts in Chapter 5 Maths NCERT. How can Vedantu help me?

If a student has any doubts, then he or she can always contact our in-house academic experts. We at Vedantu offer 24x7 query resolution services. Students can post their questions on the platform and one of our in-house experts will reply with the correct answer as soon as possible. Students can also take online classes. If there is any other way in which way we can help, you should contact us as soon as possible.

2. How can one find the sum of an arithmetic progression?

If one wants to find the sum of an arithmetic progression, then he or she must know the value of the first term, the number of terms, and the common difference that exists between each term. The following formula can be used to arrive at the final answer.


S = n / 2 (2 a + (n - 1) x d)

3. Mention the different types of progressions in mathematics.

There are three types of progressions in mathematics. These types of progressions are:

  • Arithmetic progression (AP)
  • Geometric progression (GP)
  • Harmonic progression (HP)

4. Mention some uses of arithmetic progressions.

Arithmetic progressions can be used to generalize a set of patterns that we usually find in our daily lives.

5. What kind of questions are there in NCERT Solutions for Class 10 Maths Chapter 5?

Vedantu offers solutions for the NCERT Textbook chapter-wise in a simplified manner. Chapter 5 of Class 10 Maths is Arithmetic Progressions. This chapter includes the sequence of numbers where the difference between the two consecutive numbers is the same. This chapter also finds applications in real life such as days of the week, students’ roll numbers, etc. This chapter also includes the other two types of progressions: Geometric and Harmonic.

6. Is it necessary to learn all the topics provided in NCERT Solutions for Class 10 Maths Chapter 5?

Yes, it is very important to learn all the topics and concepts provided in NCERT Solutions for Class 10 Maths Chapter 5. As this chapter forms the foundation for higher grade Mathematics, students need to know the basic formulae and their application. Vedantu offers solutions to all the exercises of chapter 5. The students can refer them online or download them for free to refer offline from the Vedantu website or the Vedantu app.

7. How many Exercises are there in Class 10 Maths Chapter 5 AP?

Chapter 5 of Class 10 Mathematics is Arithmetic progressions. This chapter includes four exercises with similar examples. The examples give you an idea to solve the exercise problems. Vedantu provides solutions to all the exercises in a step-by-step manner. One of the advantages of referring to Vedantu Solutions is that the answers are verified by the subject-matter-experts and are given in a simplified manner.

8. Is Chapter 5 Arithmetic Progression of Class 10 Maths interesting?

Yes, Chapter 5 Arithmetic Progression of Class 10 Maths is interesting. The concepts involved in this chapter are a foundation of the concepts that appear in the higher grades. Also, these have applications in real life such as roll numbers of the students in a class, months in a year, and weeks in a day. Vedantu provides the best learning while solving problems. Not only are the solutions verified, but they are also given in a simplified step-by-step manner.

9. Is Class 10 Arithmetic Progressions of Class 10 Hard?

Maths can be a tricky subject and as the board examinations approach, most of the students have fear while giving the examinations. The only solution to this is by achieving confidence and by practising different types of problems on the same concept. Vedantu offers the previous year’s question papers that the students can refer to get an idea of the marking scheme, the difficulty level of the question paper, and important questions.

10. Is arithmetic progression important?

Yes, arithmetic progression (AP) is important because it's a fundamental concept in mathematics with wide-ranging applications in various fields.

11. What is the main purpose of arithmetic?

The main purpose of arithmetic is to help us to perform fundamental mathematical operations, such as addition, subtraction, multiplication, and division. These functions are necessary for handling daily issues, handling money, calculating amounts, and recognising trends in numerical data. Whether it's counting objects, calculating the cost of groceries, or solving complex mathematical problems, arithmetic forms the basis of mathematical understanding and problem-solving skills.

12. Who is the father of arithmetic?

Brahmagupta is known as the father of arithmetic.

13. What is the full concept of arithmetic?

The full concept of arithmetic encompasses the study of basic mathematical operations involving numbers, such as addition, subtraction, multiplication, and division. It involves understanding how these operations work and how they can be applied to solve various mathematical problems. Arithmetic is fundamental to everyday life, from counting objects to managing finances and beyond. It forms the foundation of more advanced branches of mathematics and is essential for developing strong mathematical skills. In essence, arithmetic provides the tools and techniques necessary for performing calculations and solving numerical problems in a wide range of contexts.

14. What are the real life applications of arithmetic series?

Examples of Real-Life Arithmetic Sequences

  • Financial Planning

  • Population Growth

  • Seating around tables

  • Stacking cups, chairs, bowls