NCERT Solutions for 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 $1^{st}$ km is Rs. $15$.  

Taxi fare for $2^{nd}$ km is Rs. $15+8=23$.  

Taxi fare for $3^{rd}$ km is Rs. $$23+8=31$$.  

Similarly, Taxi fare for $n^{th}$ km is Rs. $$15+(n-1)8$$. 

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 ${\displaystyle \frac{1}{4}}$ of air remaining in the cylinder at a time. Hence, 

Volume after $1^{st}$ stroke is ${\displaystyle \frac{3V}{4}}$.   

Volume after $2^{nd}$ stroke is ${\displaystyle \frac{3}{4}}\left({\displaystyle \frac{3V}{4}}\right)$.   

Volume after $3^{rd}$ stroke is ${\left({\displaystyle \frac{3}{4}}\right)}^2\left({\displaystyle \frac{3V}{4}}\right)$.   

Similarly, Volume after $n^{th}$ stroke is ${\left({\displaystyle \frac{3}{4}}\right)}^{n}V$.   

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by ${\displaystyle \frac{3}{4}}$. 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 $1^{st}$ meter is Rs. $150$.  

Cost of digging for $2^{nd}$ meter is Rs. $150+50=200$.  

Cost of digging for $3^{rd}$ meter is Rs. $$200+50=250$$.  

Similarly, Cost of digging for $n^{th}$ meter is Rs. $$150+(n-1)50$$. 

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 $$\mathbf{10000}$$ is deposited at compound interest at $$\mathbf{8}\mathrm{\%}$$ per annum.

Ans: Given the principal amount is Rs.$$\mathbf{10000}$$ and the compound interest is $$\mathbf{8}\mathrm{\%}$$ per annum. Hence, 

Amount after $1^{st}$ year is Rs. $10000(1+{\displaystyle \frac{8}{100}})$.  

Amount after $2^{nd}$ year is Rs. $10000{(1+{\displaystyle \frac{8}{100}})}^2$.  

Amount after $3^{rd}$ year is Rs. $10000{(1+{\displaystyle \frac{8}{100}})}^3$.  

Similarly, Amount after $n^{th}$ year is Rs. $10000{(1+{\displaystyle \frac{8}{100}})}^n$.

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by $$(1+{\displaystyle \frac{8}{100}})$$. 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 $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(1)

Substituting $$a=10,d=10$$ in (1) we get, $a_n=10+10(n-1)=10n$ …..(2)

Therefore, from (2)

$a_1=10$, $a_2=20$, $a_3=30$ and $a_4=40$.

2. $$a=-2,d=0$$

Ans: We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(1)

Substituting $$a=-2,d=0$$ in (1) we get, $a_n=-2+0(n-1)=-2$ …..(2)

Therefore, from (2)

$a_1=-2$, $a_2=-2$, $a_3=-2$ and $a_4=-2$.

3. $$a=4,d=-3$$

Ans: We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(1)

Substituting $$a=4,d=-3$$ in (1) we get, $a_n=4-3(n-1)=7-3n$ …..(2)

Therefore, from (2)

$a_1=4$, $a_2=1$, $a_3=-2$ and $a_4=-5$.

4. $$a=-1\text{,}d=1/2$$

Ans: We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(1)

Substituting $$a=-1\text{,}d=1/2$$ in (1) we get, $a_n=-1+{\displaystyle \frac{1}{2}}(n-1)={\displaystyle \frac{n-3}{2}}$ …..(2)

Therefore, from (2)

$a_1=-1$, $a_2=-{\displaystyle \frac{1}{2}}$, $a_3=0$ and $a_4={\displaystyle \frac{1}{2}}$.

5. $$a=-1.25,d=-0.25$$

Ans: We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(1)

Substituting $$a=-1.25,d=-0.25$$ in (1) we get, $a_n=-1.25-0.25(n-1)=-1-0.25n$ …..(2)

Therefore, from (2)

$a_1=-1.25$, $a_2=-1.5$, $a_3=-1.75$ and $a_4=-2$.

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

1. $$\mathbf{3},\mathbf{1},-\mathbf{1},-\mathbf{3},...$$

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 $=$ $2^{nd}term-1^{st}term$

$\therefore$ Common difference $=$ $1-3=-2$.

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 $=$ $2^{nd}term-1^{st}term$

$\therefore$ Common difference $=$ $-1-(-5)=4$.

3. $${\displaystyle \frac{1}{3}}\text{,}{\displaystyle \frac{5}{3}}\text{,}{\displaystyle \frac{9}{3}}\text{,}{\displaystyle \frac{13}{3}},...$$

Ans: From the given AP, we can see that the first term is ${\displaystyle \frac{1}{3}}$.

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

Common difference $=$ $2^{nd}term-1^{st}term$

$\therefore$ Common difference $=$ ${\displaystyle \frac{5}{3}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{4}{3}}$.

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 $=$ $2^{nd}term-1^{st}term$

$\therefore$ Common difference $=$ $1.7-0.6=1.1$.

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

1. $$\text{2,4,8,16}...$$

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

$a_2-a_1=4-2=2$              …..(1)

$a_3-a_2=8-4=4$              …..(2)

$a_4-a_3=16-8=8$              …..(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. $$\text{2,}{\displaystyle \frac{5}{2}}\text{,3,}{\displaystyle \frac{7}{2}}\text{,}...$$

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

$a_2-a_1={\displaystyle \frac{5}{2}}-2={\displaystyle \frac{1}{2}}$              …..(1)

$a_3-a_2=3-{\displaystyle \frac{5}{2}}={\displaystyle \frac{1}{2}}$              …..(2)

$a_4-a_3={\displaystyle \frac{7}{2}}-3={\displaystyle \frac{1}{2}}$              …..(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 $2$ and common difference ${\displaystyle \frac{1}{2}}$.

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(4)

Substituting $$a=2,d={\displaystyle \frac{1}{2}}$$ in (1) we get, $a_n=2+{\displaystyle \frac{1}{2}}(n-1)={\displaystyle \frac{n+3}{2}}$ …..(5)

Therefore, from (5)

$a_5=4$, $a_6={\displaystyle \frac{9}{2}}$ and $a_7=5$.

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.

$a_2-a_1=3.2-1.2=2$              …..(1)

$a_3-a_2=5.2-3.2=2$              …..(2)

$a_4-a_3=7.2-5.2=2$              …..(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 $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(4)

Substituting $$a=1.2,d=2$$ in (1) we get, $a_n=1.2+2(n-1)=2n-0.8$ …..(5)

Therefore, from (5)

$a_5=9.2$, $a_6=11.2$ and $$a_7=13.2$$.

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.

$a_2-a_1=-6-(-10)=4$              …..(1)

$a_3-a_2=-2-(-6)=4$              …..(2)

$a_4-a_3=2-(-2)=4$              …..(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 $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(4)

Substituting $$a=-10,d=4$$ in (1) we get, $a_n=-10+4(n-1)=4n-14$ …..(5)

Therefore, from (5)

$a_5=6$, $a_6=10$ and $$a_7=14$$.

5. $$3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...$$

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

$$a_2-a_1=(3+\sqrt{2})-\left(3\right)=\sqrt{2}$$              …..(1)

$a_3-a_2=(3+2\sqrt{2})-(3+\sqrt{2})=\sqrt{2}$              …..(2)

$a_4-a_3=(3+3\sqrt{2})-(3+2\sqrt{2})=\sqrt{2}$              …..(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 $$\sqrt{2}$$. 

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(4)

Substituting $$a=3,d=\sqrt{2}$$ in (1) we get, $a_n=3+(n-1)\sqrt{2}$   …..(5)

Therefore, from (5)

$a_5=3+4\sqrt{2}$, $a_6=3+5\sqrt{2}$ and $$a_7=3+6\sqrt{2}$$.

6. $$\text{0}\text{.2,0}\text{.22,0}\text{.222,0}\text{.2222}.....$$

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

$a_2-a_1=0.22-0.2=0.02$              …..(1)

$a_3-a_2=0.222-0.22=0.002$              …..(2)

$a_4-a_3=0.2222-0.222=0.0002$              …..(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.

$a_2-a_1=-4-0=-4$              …..(1)

$a_3-a_2=-8-(-4)=-4$              …..(2)

$a_4-a_3=-12-(-8)=-4$              …..(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 $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(4)

Substituting $$a=0,d=-4$$ in (1) we get, $a_n=0-4(n-1)=4-4n$ …..(5)

Therefore, from (5)

$a_5=-16$, $a_6=-20$ and $$a_7=-24$$.

8. $-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}}....$

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

$a_2-a_1=(-{\displaystyle \frac{1}{2}})-(-{\displaystyle \frac{1}{2}})=0$              …..(1)

$a_3-a_2=(-{\displaystyle \frac{1}{2}})-(-{\displaystyle \frac{1}{2}})=0$              …..(2)

$a_4-a_3=(-{\displaystyle \frac{1}{2}})-(-{\displaystyle \frac{1}{2}})=0$              …..(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 $-{\displaystyle \frac{1}{2}}$ and common difference $0$. 

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$       …..(4)

Substituting $$a=-{\displaystyle \frac{1}{2}},d=0$$ in (1) we get, $a_n=-{\displaystyle \frac{1}{2}}+0(n-1)=-{\displaystyle \frac{1}{2}}$ …..(5)

Therefore, from (5)

$a_5=-{\displaystyle \frac{1}{2}}$, $a_6=-{\displaystyle \frac{1}{2}}$ and $$a_7=-{\displaystyle \frac{1}{2}}$$.

9. $1,3,9,27,\dots$

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

$a_2-a_1=3-1=2$         …..(1)

$a_3-a_2=9-3=6$           …..(2)

$a_4-a_3=27-9=18$           …..(3)

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

10. $a,2a,3a,4a,\dots$

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

$a_2-a_1=2a-a=a$              …..(1)

$a_3-a_2=3a-2a=a$              …..(2)

$a_4-a_3=4a-3a=a$              …..(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 $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+\left(n-1\right)d$       …..(4)

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

we get, $a_n=a+\left(n-1\right)d$....(5)

$a_5=a+(5-1)a=5a$

$a_6=a+(6-1)a=6a$

$a_7=a+(7-1)a=7a$

11. $a,a^2,a^3,a^4,\dots$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.$a_2-a_1=a^2-a=a\left(a-1\right)$              …..(1)

$a_3-a_2=a^3-a^2=a^2\left(a-1\right)$         …..(2)

$a_4-a_3=a^4-a^3=a^3\left(a-1\right)$  ……..(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. $\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},\dots$

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

$a_2-a_1=\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$     ……(1)

$a_3-a_2=\sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2}=\sqrt{2}$   ………(2)

$a_4-a_3=\sqrt{32}-\sqrt{18}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$.......(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 by$a_n=a+(n-1)d$      …..(4)

Substituting, a = $\sqrt{2}$d = $\sqrt{2}$ in (4)

we get, 

$a_5=\sqrt{2}+\left(5-1\right)\sqrt{2}=\sqrt{2}+4\sqrt{2}=5\sqrt{2}=\sqrt{50}$

$a_6=\sqrt{2}+\left(6-1\right)\sqrt{2}=\sqrt{2}+5\sqrt{2}=6\sqrt{2}=\sqrt{72}$

$a_7=\sqrt{2}+\left(7-1\right)\sqrt{2}=\sqrt{2}+6\sqrt{2}=7\sqrt{2}=\sqrt{98}$

13. $\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},\dots$

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

$a_2-a_1=\sqrt{6}-\sqrt{3}=\sqrt{3}\times \sqrt{2}-\sqrt{3}=\sqrt{3}\left(\sqrt{2}-1\right)$     ……(1)

$a_3-a_2=\sqrt{9}-\sqrt{6}=3-\sqrt{6}=\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)$   ………(2)

$a_4-a_3=\sqrt{12}-\sqrt{9}=2\sqrt{3}-3=\sqrt{3}\left(2-\sqrt{3}\right)$……….(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. $$1^2,3^2,5^2,7^2,\dots$$

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

$a_2-a_1=3^2-1^2=8$     ……(1)

$a_3-a_2=5^2-3^2=16$  ………(2)

$a_4-a_3=7^2-5^2=24$……….(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.$$1^2,5^2,7^2,73,...$$

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

$a_2-a_1=5^2-1^2=24$ ……(1)

$a_3-a_2=7^2-5^2=24$………(2)

$a_4-a_3=73-7^2=24$  ……….(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 by$a_n=a+(n-1)d$      …..(4)

Substituting a = $1^2$, d=24 in (4)

we get, 

$a_5=1^2+\left(5-1\right)24=1+\left(4\right)24=97$

$a_6=1^2+\left(6-1\right)24=1+\left(5\right)24=121$

$a_7=1^2+\left(7-1\right)24=1+\left(6\right)24=145$

Exercise 5.2

1. Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $$a_n$$ the $$n^{th}$$ term of the A.P.

(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 $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

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

$a_n=7+(8-1)3$

$\Rightarrow a_n=7+21$

$\therefore a_n=28$

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

Given, the $n^{th}$ term, $$a_n=0$$ …..(2)

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

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

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

$0=-18+(10-1)d$

$\Rightarrow 18=9d$

$\therefore d=2$

(iii). Ans: Given, the $n^{th}$ term, $$a_n=-5$$ ….. (1)

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

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

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

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

$-5=a+(18-1)(-3)$

$\Rightarrow -5=a-51$

$\therefore a_n=46$

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

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

Given, the $n^{th}$ term, $$a_n=3.6$$ …..(3)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

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

$3.6=-18.9+(n-1)\left(2.5\right)$

$\Rightarrow 22.5=(n-1)\left(2.5\right)$

$\Rightarrow 9=(n-1)$

$\therefore a_n=10$

(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 $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

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

$a_n=3.5+(105-1)\left(0\right)$

$\therefore a_n=3.5$

2. Choose the correct choice in the following and justify

(i). $${30}^{th}$$ 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, $$d=7-10=-3$$ …..(2)

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

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $a_n=10+(30-1)(-3)$

$\Rightarrow a_n=10-87$

$\therefore a_n=-77$

(ii). $${11}^{th}$$ term of the A.P $$-3,-{\displaystyle \frac{1}{2}},2,...,$$ is

1. $$28$$

2. $$22$$ 

3. $$38$$

4. $$48{\displaystyle \frac{1}{2}}$$

Ans: B. $22$ 

Given, the first Term, $a=-3$  ….. (1)

Given, the common Difference, $$d=-{\displaystyle \frac{1}{2}}-(-3)={\displaystyle \frac{5}{2}}$$ …..(2)

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

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $a_n=-3+{\displaystyle \frac{5}{2}}(11-1)$

$\Rightarrow a_n=-3+25$

$\therefore a_n=22$

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

(i). $$2,\mathrm{\_}\mathrm{\_},26$$ 

Ans: Given, first term $a=2$ …..(1)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     

Substituting the values from (1) we get, $a_n=2+(n-1)d$  …..(2)

Given, third term $a_3=26$. From (2) we get, 

$26=2+(3-1)d$

$\Rightarrow 26=2+2d$ 

$\therefore d=12$  ….(3)

From (1), (2) and (3) we get for $n=2$ 

$a_2=2+(2-1)\left(12\right)$

$\therefore a_2=14$

$\therefore$ The sequence is $$2,14,26$$.

(ii). $$\mathrm{\_}\mathrm{\_},13,\mathrm{\_}\mathrm{\_},3$$

Ans: Given, second term $a_2=13$ …..(1)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$ …..(2)

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

Given, fourth term $a_4=3$. From (2) we get, 

$3=a+3d$  …..(4)

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

$3-13=(a+3d)-(a+d)$

$\Rightarrow -10=2d$ 

$\therefore d=-5$  ….(5)

From (3) and (5) we get 

$13=a-5$

$\Rightarrow a=18$ ……(6)

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

$a_n=18-5(n-1)$   …..(7)

First term, $a=18$ and third term $a_3=8$

$\therefore$ The sequence is $$18,13,8,3$$.

(iii). $5,\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},9{\displaystyle \frac{1}{2}}$

Ans: Given, first term $a=5$ …..(1)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$ …..(2)

Substituting the values from (1) in (2) we get, $a_n=5+(n-1)d$  …..(3)

Given, fourth term $a_4=9{\displaystyle \frac{1}{2}}$. From (3) we get, 

$9{\displaystyle \frac{1}{2}}=5+(4-1)d$ 

$\Rightarrow 9{\displaystyle \frac{1}{2}}=5+3d$ 

$\therefore d={\displaystyle \frac{3}{2}}$  ….(4)

From (3) and (4) we get 

$a_n=5+{\displaystyle \frac{3}{2}}(n-1)$ ……(5)

Second term, $a_2={\displaystyle \frac{13}{2}}$ and third term $a_3=8$

$\therefore$ The sequence is $5,{\displaystyle \frac{13}{2}},8,9{\displaystyle \frac{1}{2}}$.

(iv). $-4,\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},6$

Ans: Given, first term $a=-4$ …..(1)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$ …..(2)

Substituting the values from (1) in (2) we get, $a_n=-4+(n-1)d$  …..(3)

Given, sixth term $a_6=6$. From (3) we get, 

$6=-4+(6-1)d$ 

$\Rightarrow 6=-4+5d$ 

$\therefore d=2$  ….(4)

From (3) and (4) we get 

$a_n=-4+2(n-1)$ ……(5)

Second term $a_2=-2$, third term $a_3=0$, fourth term $a_4=2$ and fifth term $a_5=4$.

$\therefore$ The sequence is $-4,-2,0,2,4,6$

(v). $\mathrm{\_}\mathrm{\_},38,\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},-22$

Ans: Given, second term $a_2=38$ …..(1)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$ …..(2)

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

Given, sixth term $a_6=-22$. From (2) we get, 

$-22=a+5d$  …..(4)

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

$-22-38=(a+5d)-(a+d)$

$\Rightarrow -60=4d$ 

$\therefore d=-15$  ….(5)

From (3) and (5) we get 

$38=a-15$

$\Rightarrow a=53$ ……(6)

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

$a_n=53-15(n-1)$   …..(7)

First term, $a=53$, second term $a_3=23$, third term $a_3=8$ and fourth term $a_4=-7$

$\therefore$ 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, $$d=8-3=5$$ …..(2)

Given, the $n^{th}$ term, $$a_n=78$$ …..(3)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$     …..(4)

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

$78=3+5(n-1)$

$\Rightarrow 75=5(n-1)$

$\Rightarrow 15=(n-1)$