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

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Last updated date: 27th Sep 2023
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Competitive Exams after 12th Science

Difference Between the NCERT Exemplar and the NCERT Textbook

Choose the correct answer from the given four options:

Sample Question 1 : The 10th term of the AP: 5, 8, 11, 14, ... is

(A) 32 (B) 35 (C) 38 (D) 185

Ans: Given AP is 5, 8, 11, 14,... Here, $a = 5$ and $d = 3$

Now, since

$\Rightarrow {a_n} = a + \left( {n - 1} \right)d$

So we have

$\Rightarrow {a_{10}} = 5 + \left( {10 - 1} \right)\left( 3 \right)$

Hence,

$\Rightarrow {a_{10}} = 32$

So the correct answer is (A).

Sample Question 2 : In an AP if a = –7.2, d = 3.6, ${a_n} = 7.2$ , then n is

(A) 1 (B) 3 (C) 4 (D) 5

Ans: Given that, in an AP we have $a = - 7.2$ and $d = 3.6$

Now if ${a_n} = 7.2$ then we have

$\Rightarrow 7.2 = - 7.2 + \left( {n - 1} \right)\left( {3.6} \right)$

That gives us,

$\Rightarrow \left( {n - 1} \right) = \dfrac{{2 \times 7.2}}{{3.6}}$

i.e.

$\Rightarrow n = 5$

Hence, the correct answer is (D).

Choose the correct answer from the given four options in the following questions:

1. In an A.P., if d = –4, n = 7, ${a_n} = 4$ , then a is

(a) 6

(b) 7

(c) 20

(d) 28

Ans: Since ${a_n} = a + \left( {n - 1} \right)d$

Hence, that gives us,

$\Rightarrow 4 = a + \left( {7 - 1} \right)\left( { - 4} \right)$

i.e.

$\Rightarrow a = 28$

Hence, the correct answer is (d).

2. In an A.P., if a = 3.5, d = 0, n = 101, then ${a_n}$ will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Ans: Since ${a_n} = a + \left( {n - 1} \right)d$

Hence, that gives us,

$\Rightarrow {a_n} = 3.5 + \left( {101 - 1} \right)\left( 0 \right)$

i.e.

$\Rightarrow {a_n} = 3.5$

Hence, the correct answer is (b).

3. The list of numbers  –10, –6, –2, 2, … is

(a) an A.P. with d = –16

(b) an A.P. with d = 4

(c) an A.P. with d = –4

(d) not an A.P.

Ans: The given list of numbers  –10, –6, –2, 2, … is an A.P. where $a = - 10$ and $d = 4$

Hence, the correct answer is (b).

4. The 11th term of the A.P. $- 5,\dfrac{{ - 5}}{2},0,\dfrac{5}{2},...$ is

(a) –20

(b) 20

(c) –30

(d) 30

Ans: In the given sequence, we have $a = - 5$ and $d = \dfrac{5}{2}$

Hence, ${a_{11}}$ will be given by,

$\Rightarrow {a_{11}} = - 5 + \left( {11 - 1} \right)\dfrac{5}{2}$

i..e

$\Rightarrow {a_{11}} = 20$

Hence, the correct answer is (b).

5. The first four terms of an A.P., whose first term is –2 and the common difference is (– 2), are

(a) –2, 0, 2, 4

(b) –2, 4, –8, 16

(c) –2, –4, –6, –8

(d) –2, –4, –8, –16

Ans: Given that $a = - 2$ and $d = - 2$

Therefore, the given A.P is $- 2, - 4, - 6, - 8, - 10,...$

Hence, the correct answer is (c).

6. The 21st term of the A.P. whose first two terms are –3 and 4 is

(a) 17

(b) 137

(c) 143

(d) –143

Ans: Given that ${a_1} = - 3$ and ${a_2} = 4$ hence, $d = {a_2} - {a_1} = 4 - \left( { - 3} \right)$ i.e. $d = 7$

Hence,

$\Rightarrow {a_{21}} = - 3 + \left( {21 - 1} \right)7$

i.e.

$\Rightarrow {a_{21}} = 137$

Hence, the correct answer is (b).

7. If the 2nd term of an A.P. is 13 and 5th term is 25, what is its 7th term ?

(a) 30

(b) 33

(c) 37

(d) 38

Ans: Given that ${a_2} = a + d = 13$ and ${a_5} = a + 4d = 25$ hence we have

$\Rightarrow a + 4d - a - d = 25 - 13 = 12$

That gives

$\Rightarrow 3d = 12$

i.e.

$\Rightarrow d = 4$

So ${a_1} = 13 - 4 = 9$

And hence,

$\Rightarrow {a_7} = 9 + \left( {7 - 1} \right)4$

i.e.

$\Rightarrow {a_7} = 33$

Hence, the correct answer is (b).

8. Which term of the A.P. : 21, 42, 63, 84, … is 210 ?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Ans: Given that $a = 21$ , ${a_n} = 210$ and $d = 42 - 21 = 21$

Hence,

$\Rightarrow 210 = 21 + \left( {n - 1} \right)21$

i.e.

$\Rightarrow n = 10$

Hence, the correct answer is (c).

9. If the common difference of an A.P. is 5, then what is ${a_{18}} - {a_{13}}$ ?

(a) 5

(b) 20

(c) 25

(d) 30

Ans: Given that, $d = 5$

Now, ${a_{18}} = a + 17d$ and ${a_{13}} = a + 12d$

Hence,

$\Rightarrow {a_{18}} - {a_{13}} = a + 17d - a - 12d$

i.e.

$\Rightarrow {a_{18}} - {a_{13}} = 5d$

That is,

$\Rightarrow {a_{18}} - {a_{13}} = 25$

Hence, the correct answer is (c).

10. What is the common difference of an A.P. in which ${a_{18}} - {a_{14}} = 32$ ?

(a) 8

(b) –8

(c) –4

(d) 4

Ans: Given that, ${a_{18}} - {a_{14}} = 32$ i.e. $a + 17d - a - 13d = 32$

Hence we have

$\Rightarrow a + 17d - a - 13d = 32$

Or,

$\Rightarrow 4d = 32$

So the common difference is $d = 8$

Hence, the correct answer is (a).

11. Two APs have the same common difference. The 1st term of one of these is –1 and that of the other is –8. Then the difference between their 4th terms is

(a) –1

(b) –8

(c) 7

(d) –9

Ans: Two APs have the same common difference, say d.

The fourth term of first AP will be $- 1 + 3d$ and that of the second AP will be $- 8 + 3d$

Hence, their difference is $- 1 + 3d - \left( { - 8 + 3d} \right) = 7$

Hence, the correct answer is (c).

12. If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then its 18th term will be

(a) 7

(b) 11

(c) 18

(d) 0

Ans: According to the question, we have

$\Rightarrow 7\left( {a + 6d} \right) = 11\left( {a + 10d} \right)$

That gives

$\Rightarrow 4a + 68d = 0$

i.e.

$\Rightarrow 4\left( {a + 17d} \right) = 0$

Or,

$\Rightarrow a + 17d = 0$

Hence, the correct answer is (d).

13. The 4th term from the end of the A.P. –11, –8, –5, …, 49 is

(a) 37

(b) 40

(c) 43

(d) 58

Ans: Given AP is –11, –8, –5, …, 49

Consider the reverse of this AP, i.e. where $a = 49$ and $d = - 3$

Now, we have

$\Rightarrow {a_4} = a + 3d$

i.e.

$\Rightarrow {a_4} = 49 + 3\left( { - 3} \right)$

Or,

$\Rightarrow {a_4} = 40$

Hence, the correct answer is (b).

14. The famous mathematician associated with finding the sum of the first 100 natural numbers is

(a) Pythagoras

(b) Newton

(c) Gauss

(d) Euclid

Ans: Gauss was the first mathematician associated with finding the sum of the first 100 natural numbers.

The sum of first natural numbers can be found considering the AP 1,2,3,...100

Here, we have $a = d = 1$ and $n = 100$

The sum of an AP is given by the formula

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

That gives us,

$\Rightarrow {S_{100}} = \dfrac{{100}}{2}\left[ {2\left( 1 \right) + \left( {100 - 1} \right) \cdot 1} \right]$

Or,

$\Rightarrow {S_{100}} = 50\left[ {2 + 99} \right]$

i.e.

$\Rightarrow {S_{100}} = 5050$

Hence, the correct answer is (c).

15. If the first term of an A.P. is –5 and the common difference is 2, then the sum of first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

Ans: Given that $a = - 5$ and $d = 2$ and $n = 6$

Hence,

$\Rightarrow {S_6} = \dfrac{6}{2}\left[ {2\left( { - 5} \right) + \left( {6 - 1} \right) \cdot 2} \right]$

Or,

$\Rightarrow {S_6} = 3\left[ { - 10 + 10} \right]$

i.e.

$\Rightarrow {S_6} = 0$

Hence, the correct answer is (a).

16. The sum of first 16 terms of the A.P. 10, 6, 2, … is

(a) –320

(b) 320

(c) –352

(d) –400

Ans: Given that $a = 10$ and $d = 6 - 10 = - 4$ and $n = 16$

Hence,

$\Rightarrow {S_{16}} = \dfrac{{16}}{2}\left[ {2\left( {10} \right) + \left( {16 - 1} \right) \cdot \left( { - 4} \right)} \right]$

Or,

$\Rightarrow {S_{16}} = 8\left[ {20 - 60} \right]$

i.e.

$\Rightarrow {S_{16}} = - 320$

Hence, the correct answer is (a).

17. In an A.P., if a = 1, ${a_n} = 20$ and ${S_n} = 399$ then n is

(a) 19

(b) 21

(c) 38

(d) 42

Ans: Given that, $a = 1$ , ${a_n} = 20$ and ${S_n} = 399$

Therefore, we can write

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right) \cdot d} \right]$

i.e.

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$

Or,

$\Rightarrow 399 = \dfrac{n}{2}\left[ {1 + 20} \right]$

That gives,

$\Rightarrow n = \dfrac{{798}}{{21}} = 38$

Hence, the correct answer is (c).

18. The sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

Ans: The sum of first five multiples of 3 can be found considering $a = 3$ , $d = 3$ and $n = 5$

Hence,

$\Rightarrow {S_5} = \dfrac{5}{2}\left[ {2\left( 3 \right) + \left( {5 - 1} \right) \cdot 3} \right]$

i.e.

$\Rightarrow {S_5} = \dfrac{5}{2}\left[ {18} \right]$

Or,

$\Rightarrow {S_5} = 45$

Hence, the correct answer is (a).

Short Answer Questions with Reasoning

Sample Question 1: In the AP: 10, 5, 0, –5, ... the common difference d is equal to 5. Justify whether the above statement is true or false.

Ans: Given that,  an AP as 10,5,0,-5,...

Here we have the first term, i.e. $a = 10$ and the common difference as $d = 5 - 10$

Hence, we have $d = - 5$

So the given statement is false that the common difference d is equal to 5 because it is -5.

Sample Question 2 : Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amounts at the end of first year, second year, third year, ..., form an AP. Justify your answer.

Ans: Given that, Divya deposited Rs 1000 at compound interest at the rate of 10% per annum.

Here, the amounts at the end of each year can be given as $1100,1210,1331,...$

Since the common difference in the above sequence is not unique and varies between each interval, therefore the given sequence does not form an AP.

Sample Question 3: The nth term of an AP cannot be ${n^2} + 1$ . Justify your answer.

Ans: Consider, ${a_n} = {n^2} + 1$

Then we can write,

$\Rightarrow {a_1} = {\left( 1 \right)^2} + 1 = 2$

And,

$\Rightarrow {a_2} = {\left( 2 \right)^2} + 1 = 5$

Also,

$\Rightarrow {a_3} = {\left( 3 \right)^2} + 1 = 10$

Here, ${a_2} - {a_1} = 3$ but ${a_3} - {a_2} = 5$

Since the common difference is not unique and varies between each consecutive terms, hence ${a_n} = {n^2} + 1$ can be the nth term of an AP.

EXERCISE 5.2

1. Which of the following form an A.P. ? Justify your answer.

(i) –1, –1, –1, –1, …

Ans: Here, all the terms are equal to each other i.e. the common difference is zero everywhere. So, the sequence –1, –1, –1, –1,… forms an AP.

(ii) 0, 2, 0, 2, …

Ans: Here, the given sequence is 0, 2, 0, 2,… where the common difference is not unique and varying between 2 and -2. Hence, the sequence 0, 2, 0, 2,… does not form an AP.

(iii) 1, 1, 2, 2, 3, 3, …

Ans: Here, the given sequence is 1, 1, 2, 2, 3, 3,… where the common difference is not unique and varying between 0 and 1. Hence, the sequence 1, 1, 2, 2, 3, 3,… does not form an AP.

(iv) 11, 22, 33, …

Ans: Here, the given sequence is 11, 22, 33,… where the common difference is unique and constant everywhere i.e. 11 . Hence, the sequence 11, 22, 33,… forms an AP.

(v) $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},...$

Ans: Here, the given sequence is $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},...$ where the common difference is not unique and varying everywhere. We have $\dfrac{1}{3} - \dfrac{1}{2} \ne \dfrac{1}{4} - \dfrac{1}{3}$ i.e. $- \dfrac{1}{6} \ne - \dfrac{1}{{12}}$

Hence, the sequence $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},...$ does not form an AP.

(vi) $2,{2^2},{2^3},{2^4},...$

Ans: Here, the given sequence is $2,{2^2},{2^3},{2^4},...$ where the common difference is not unique and varying everywhere. We have $4 - 2 \ne 8 - 4 \ne 16 - 8$ i.e. $2 \ne 4 \ne 8$

Hence, the sequence $2,{2^2},{2^3},{2^4},...$ does not form an AP.

(vii) $\sqrt 3 ,\sqrt {12} ,\sqrt {27} ,\sqrt {48} ,...$

Ans: Here, the given sequence is $\sqrt 3 ,\sqrt {12} ,\sqrt {27} ,\sqrt {48} ,...$ where the common difference is not unique and varies everywhere. We have $\sqrt {12} - \sqrt 3 \ne \sqrt {27} - \sqrt {12} \ne \sqrt {48} - \sqrt {27}$

Hence, the sequence $\sqrt 3 ,\sqrt {12} ,\sqrt {27} ,\sqrt {48} ,...$ does not form an AP.

2. Justify whether it is true to say that $- 1,\dfrac{{ - 3}}{2}, - 2,\dfrac{5}{2},...$ forms an A.P. as ${a_2} - {a_1} = {a_3} - {a_2}$ .

Ans: Given sequence is $- 1,\dfrac{{ - 3}}{2}, - 2,\dfrac{5}{2},...$

Here, we have $\dfrac{{ - 3}}{2} - \left( { - 1} \right) = \dfrac{{ - 1}}{2}$ and $- 2 - \left( {\dfrac{{ - 3}}{2}} \right). = \dfrac{{ - 1}}{2}$

But, $\dfrac{5}{2} - \left( { - 2} \right) = \dfrac{9}{2}$

Hence, the common difference is not constant everywhere so the given sequence is not an AP.

3. For the A.P. –3, –7, –11, …, can we find directly ${a_{30}} - {a_{20}}$ without actually finding ${a_{30}}$ and ${a_{20}}$ ? Give reasons for your answer.

Ans: Given AP is –3, –7, –11,... Here $d = - 7 - \left( { - 3} \right) = - 4$

Now,

$\Rightarrow {a_{30}} - {a_{20}} = a + 29d - \left( {a + 19d} \right)$

That is,

$\Rightarrow {a_{30}} - {a_{20}} = 10d$

Hence,

$\Rightarrow {a_{30}} - {a_{20}} = - 40$

4. Two A.P.s have the same common difference. The first term of one A.P. is 2, and that of the other is 7. The difference between their 10th terms is same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why ?

Ans: Given that, two A.P.s have the same common difference, say d.

Now their first terms are $a = 2$ and $b = 7$ respectively.

The difference between their 10th terms is,

$\Rightarrow {a_{10}} - {b_{10}} = a + 9d - \left( {b + 9d} \right)$

i.e.

$\Rightarrow {a_{10}} - {b_{10}} = a - b = - 5$

Also, the difference between their 21st terms is,

$\Rightarrow {a_{21}} - {b_{21}} = a + 20d - \left( {b + 20d} \right)$

i.e.

$\Rightarrow {a_{21}} - {b_{21}} = a - b = - 5$

Now, the difference between any two corresponding terms is,

$\Rightarrow {a_n} - {b_n} = a + \left( {n - 1} \right)d - \left[ {b + \left( {n - 1} \right)d} \right]$

i.e.

$\Rightarrow {a_n} - {b_n} = a - b = - 5$

Therefore, the difference between their 10th terms is same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.

5. Is 0 a term of the A.P. 31, 28, 25, … ∵ Justify your answer.

Ans: In the given AP, we have $a = 31$ , $d = 28 - 31 = - 3$

Now if ${a_n} = 0$ , then we have

$\Rightarrow {a_n} = a + \left( {n - 1} \right)d$

That gives

$\Rightarrow 0 = 31 + \left( {n - 1} \right)\left( { - 3} \right)$

Or,

$\Rightarrow \left( {n - 1} \right) = - \dfrac{{31}}{{\left( { - 3} \right)}}$

i.e.

$\Rightarrow n = \dfrac{{31 + 3}}{3}$

Hence,

$\Rightarrow n = \dfrac{{34}}{3} \notin \mathbb{N}$

Therefore, 0 is not a term of the given AP.

6.The taxi fare after each km, when the fare is Rs 15 for the first km and Rs 8 for each additional km, does not form an A.P., as the total fare (in Rs) after each km is 15, 8, 8, 8, … Is the statement true ∵ Give reasons.

Ans: Yes, the sequence of the total fare does not form an AP. This statement is true because the total fare that is given as 15, 8, 8, 8, … is not the total fare of 1,2,3,4,5,.. kilometres respectively.

Here, we have $a = 15$  and ${d_1} = 8 - 15 = - 7$

But, ${d_1} = - 7 \ne {d_2} = {d_3} = {d_4} = ... = {d_n} = 0$

The given sequence is in the form of $a,d,d,d,d...$ instead of $a,a + d,a + 2d,a + 3d,...$

Hence, the given terms do not form an AP.

7. In which of the following situations do the lists of numbers involved form an A.P.? Give reasons for your answers.

(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.

(ii) The fee charged every month by a school from classes I to XII, when the monthly fee for class I is Rs 250, and it increase by Rs 50 for the next higher class.

(iii) The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.

(iv) The number of bacteria in a certain food item after each second, when they double in every second.

Ans: (i) If the monthly fee is Rs 400, then the fee charged from a student every month by a school for the whole session, is given by $400,400,400,....$

Here, $a = 400$ and $d = 0$ .

Therefore, it forms an AP.

(ii) If the monthly fee for class I is Rs 250, and it increases by Rs 50 for the next higher class.

Then the fee charged every month by a school from classes I to XII, is given by 250,300,350,400,...

Here, $a = 250$ and $d = 50$ .

Therefore, it forms an AP.

(iii) If Rs 1000 is deposited at simple interest of 10% per annum, then the amount of money in the account of Varun at the end of every year is given by, 1000,1100,1200,1300,...

Here $a = 1000$ and $d = 100$

Therefore, it does not form an AP.

(iv) If the number of bacteria double in every second in a certain food item, then the number of bacteria after each second is given by a,2a,4a,8a,16a,...

Here the common difference is not unique and varies at every interval.

Therefore, it does not form an AP.

8. Justify whether it is true to say that the following are the nth terms of an A.P.

(i) $2n – 3$

Ans: We have ${a_n} = 2n - 3$

Now, we can write ${a_1} = 2 - 3 = - 1$ , ${a_2} = 4 - 3 = 1$ , ${a_3} = 6 - 3 = 3$ and so on.

Here, we have $a = - 1$ and $d = 2$

Hence 2n – 3 is an nth term of an A.P.

(ii) $3{n^2} + 5$

Ans: We have ${a_n} = 3{n^2} + 5$

Now, we can write ${a_1} = 3 + 5 = 8$ , ${a_2} = 12 + 5 = 17$ , ${a_3} = 27 + 5 = 32$ and so on.

Here the common difference is not unique and varies at every interval.

Hence $3{n^2} + 5$ is not an nth term of an A.P.

(iii) $1 + n + {n^2}$

Ans: We have ${a_n} = 1 + n + {n^2}$

Now, we can write ${a_1} = 1 + 1 + 1 = 3$ , ${a_2} = 1 + 2 + 4 = 7$ , ${a_3} = 1 + 3 + 9 = 13$ and so on.

Here the common difference is not unique and varies at every interval.

Hence $1 + n + {n^2}$ is not an nth term of an A.P.

Sample Question 1 : If the numbers n – 2, 4n – 1 and 5n + 2 are in AP, find the value of n.

Ans: Given that, three numbers $n - 2$ , $4n - 1$ and $5n + 2$ are in AP.

Now we have

$\Rightarrow \left( {4n - 1} \right) - \left( {n - 2} \right) = 3n + 1$

And,

$\Rightarrow \left( {5n + 2} \right) - \left( {4n - 1} \right) = n + 3$

Now if they are if then the common difference must be constant.

Hence, we can write

$\Rightarrow 3n + 1 = n + 3$

That gives us,

$\Rightarrow 2n = 2$

Therefore,

$\Rightarrow n = 1$

Sample Question 2 : Find the value of the middle most term (s) of the AP : –11, –7, –3,..., 49.

Ans: Given AP is –11, –7, –3,..., 49.

Here we have $a = - 11$ , $d = 4$ and ${a_n} = 49$

So we can write,

$\Rightarrow {a_n} = a + \left( {n - 1} \right)d$

Or,

$\Rightarrow 49 = - 11 + \left( {n - 1} \right)\left( 4 \right)$

That gives us,

$\Rightarrow n = 16$

Since n is an even number, so there are two middle terms i.e. the $\dfrac{n}{2}th$ and $\left( {\dfrac{n}{2} + 1} \right)th$ terms.

Hence the middle most terms are ${a_8}$ and ${a_9}$ .

Now,

$\Rightarrow {a_8} = - 11 + 7\left( 4 \right)$

i.e.

$\Rightarrow {a_8} = 17$

Ans,

$\Rightarrow {a_9} = - 11 + 8\left( 4 \right)$

i.e.

$\Rightarrow {a_9} = 21$

Sample Question 3: The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, find the AP.

Ans: According to the given question, we have ${a_1} + {a_2} + {a_3} = 33$

Let ${a_1} = a - d$ , ${a_2} = a$ and ${a_3} = a + d$

Now that can be written as,

$\Rightarrow a - d + a + a + d = 33$

i.e.

$\Rightarrow 3a = 33$

Hence,

$\Rightarrow a = 11$

Now it is also given that ${a_1}{a_3} = {a_2} + 29$

That gives us,

$\Rightarrow \left( {a - d} \right)\left( {a + d} \right) = a + 29$

That is,

$\Rightarrow \left( {11 - d} \right)\left( {11 + d} \right) = 11 + 29$

Or,

$\Rightarrow {\left( {11} \right)^2} - {d^2} = 40$

That can be written as,

$\Rightarrow {d^2} = 121 - 40$

Hence,

$\Rightarrow {d^2} = 81$

i.e.

$\Rightarrow d = \pm 9$

Therefore, the required AP is either $2,11,20,29,38,...$ or $20,11,2, - 7,...$

EXERCISE 5.3

1. Match the A.P.s given in column A with suitable common differences given in column B.

 Column A Column B (A1 ) 2, –2, –6, –10,… (A2 ) a = –18, n = 10, ${a_n} = 0$(A3 ) a = 0, ${a_{10}} = 6$(A4 ) ${a_2} = 13$ , ${a_4} = 3$ (B1 ) 2/3(B2 ) –5(B3 ) 4(B4 ) –4(B5 ) 2(B6 ) 1/2(B7 ) 5

Ans: (1) We have 2, –2, –6, –10,...

Here, we have

$\Rightarrow d = - 2 - 2 = - 4$

(2) We have a = –18, n = 10, ${a_n} = 0$

Now, we can write

$\Rightarrow 0 = - 18 + \left( {10 - 1} \right)d$

Hence,

$\Rightarrow d = 2$

(3) We have a = 0, ${a_{10}} = 6$

Hence, we can write

$\Rightarrow 6 = 0 + \left( {10 - 1} \right)d$

i.e.

$\Rightarrow d = \dfrac{2}{3}$

(4) We have${a_2} = 13$ , ${a_4} = 3$

Hence, we can write

$\Rightarrow a + 3d - \left( {a + d} \right) = 13 - 3$

Or,

$\Rightarrow 2d = 10$

i.e.

$\Rightarrow d = 5$

2. Verify that each of the following is an A.P. and then write its next three terms.

(i) $0,\dfrac{1}{4},\dfrac{1}{2},\dfrac{3}{4},...$

Ans: Here, we have $a = 0$ and ${d_1} = \dfrac{1}{4} - 0 = \dfrac{1}{4}$ , ${d_2} = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}$ , ${d_3} = \dfrac{3}{4} - \dfrac{1}{2} = \dfrac{1}{4}$

Therefore, ${d_{}} = \dfrac{1}{4}$ so the given terms form an AP.

Its next three terms will be given by $0,\dfrac{1}{4},\dfrac{1}{2},\dfrac{3}{4},1,\dfrac{5}{4},\dfrac{3}{2},\dfrac{7}{4},\dfrac{8}{4},...$

(ii) $5,\dfrac{{14}}{3},\dfrac{{13}}{3},4,...$

Ans: Here, we have $a = 5$ and ${d_1} = \dfrac{{14}}{3} - 5 = - \dfrac{1}{3}$ , ${d_2} = \dfrac{{13}}{3} - \dfrac{{14}}{3} = - \dfrac{1}{3}$ , ${d_3} = 4 - \dfrac{{13}}{3} = - \dfrac{1}{3}$

Therefore, $d = - \dfrac{1}{3}$ so the given terms form an AP.

Its next three terms will be given by $5,\dfrac{{14}}{3},\dfrac{{13}}{3},4,\dfrac{{11}}{3},\dfrac{{10}}{3},3,\dfrac{8}{3},...$

(iii) $\sqrt 3 ,2\sqrt 3 ,3\sqrt 3 ,...$

Ans: Here, we have $a = \sqrt 3$ and ${d_1} = 2\sqrt 3 - \sqrt 3 = \sqrt 3$ , ${d_2} = 3\sqrt 3 - 2\sqrt 3 = \sqrt 3$ , ${d_3} = 4\sqrt 3 - 3\sqrt 3 = \sqrt 3$

Therefore, $d = \sqrt 3$ so the given terms form an AP.

Its next three terms will be given by $\sqrt 3 ,2\sqrt 3 ,3\sqrt 3 ,4\sqrt 3 ,5\sqrt 3 ,6\sqrt 3 ,7\sqrt 3 ,...$

(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …

Ans: Here, we have ${a_1} = a + b$ and ${d_1} = \left( {a + 1} \right) + b - a - b = 1$ , ${d_2} = \left( {a + 1} \right) + \left( {b + 1} \right) - \left( {a + 1} \right) - b = 1$

Therefore, $d = 1$ so the given terms form an AP.

Its next three terms will be given by $\left( {a + b} \right),\left( {a + 1} \right) + b,\left( {a + 1} \right) + \left( {b + 1} \right),\left( {a + 2} \right) + \left( {b + 1} \right),\left( {a + 2} \right) + \left( {b + 2} \right),\left( {a + 3} \right) + \left( {b + 1} \right),...$

(v) a, 2a + 1, 3a + 2, 4a + 3, …

Ans: Here, we have ${a_1} = a$ and ${d_1} = \left( {2a + 1} \right) - a = a + 1$ , ${d_2} = \left( {3a + 2} \right) - \left( {2a + 1} \right) = a + 1$ , ${d_3} = \left( {4a + 3} \right) - \left( {3a + 2} \right) = a + 1$

Therefore, $d = a + 1$ so the given terms form an AP.

Its next three terms will be given by $a,\left( {2a + 1} \right),\left( {3a + 2} \right),\left( {4a + 3} \right),\left( {5a + 4} \right),\left( {6a + 5} \right),\left( {7a + 6} \right),...$

3. Write the first three terms of the A.P.s when a and d are as given below.

(i) $a = \dfrac{1}{2},d = \dfrac{{ - 1}}{6}$

Ans: Here, $a = \dfrac{1}{2},d = \dfrac{{ - 1}}{6}$

Hence, the AP is given by $\dfrac{1}{2},\dfrac{1}{2} - \dfrac{1}{6},\dfrac{1}{2} - \dfrac{2}{6},\dfrac{1}{2} - \dfrac{3}{6},\dfrac{1}{2} - \dfrac{4}{6},..$

That is, $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{6},0,\dfrac{{ - 1}}{6},..$

(ii) $a = - 5,d = - 3$

Ans: Here, $a = - 5,d = - 3$

Hence, the AP is given by $- 5, - 5 - 3, - 5 - 6, - 5 - 9, - 5 - 12,...$

That is, $- 5, - 8, - 11, - 14, - 17,...$

(iii) $a = \sqrt 2 ,d = \dfrac{1}{{\sqrt 2 }}$

Ans: Here, we have $a = \sqrt 2$ and $d = \dfrac{1}{{\sqrt 2 }}$

Hence, the terms of this AP are given as

$\Rightarrow a,a + d,a + 2d,a + 3d,...$

That gives us,

$\Rightarrow \sqrt 2 ,\sqrt 2 + \dfrac{1}{{\sqrt 2 }},\sqrt 2 + \dfrac{2}{{\sqrt 2 }},\sqrt 2 + \dfrac{3}{{\sqrt 2 }},...$

Or,

$\Rightarrow \sqrt 2 ,\dfrac{{3\sqrt 2 }}{2},2\sqrt 2 ,\dfrac{5}{{\sqrt 2 }},3\sqrt {2,} ...$

Therefore, the given AP is $\sqrt 2 ,\dfrac{{3\sqrt 2 }}{2},2\sqrt 2 ,\dfrac{5}{{\sqrt 2 }},3\sqrt {2,} ...$

4. Find a, b and c such that the following numbers are in A.P.: a, 7, b, 23, c.

Ans: Given terms are, $a,7,b,23,c,...$

For the given terms to from an AP, we must have $7 - a = b - 7 = 23 - b = c - 23$

That gives us, $a + b = 14,2b = 30,b + c = 46$

Hence, $b = 15$ so $a = - 1$ and $c = 31$

So the given AP is $- 1,7,15,23,31,...$

5. Determine the A.P. whose 5th term is 19 and the difference of the 8th term from the 13th term is 20.

Ans: Given that ${a_5} = 19$ and ${a_{13}} - {a_8} = 20$

That gives us, $a + 4d = 19$ and $a + 12d - a - 7d = 20$or $5d = 20$

Hence, $d = 4$

So we have $a = 19 - 4\left( 4 \right) = 3$

Hence, the given AP is $3,7,11,15,19,...$

6. The 26th, 11th and the last term of an A.P. are 0, 3 and $- \dfrac{1}{5}$ respectively. Find the common difference and the number of terms.

Ans: Given that ${a_{26}} = 0,{a_{11}} = 3,{a_n} = - \dfrac{1}{5}$

That gives us ${a_{26}} - {a_{11}} = a + 25d - a - 10d = 0 - 3$ i.e. $15d = - 3$

Hence, $d = - \dfrac{1}{5}$ and $a + 25\left( { - \dfrac{1}{5}} \right) = 0$ i.e. $a = 5$

Now if $- \dfrac{1}{5} = 5 + \left( {n - 1} \right)\left( { - \dfrac{1}{5}} \right)$

Then we have

$\Rightarrow 26 = \left( {n - 1} \right)$

i.e.

$\Rightarrow n = 27$

7. The sum of the 5th and the 7th terms of an A.P. is 52, and the 10th term is 46. Find the A.P.

Ans: Given that ${a_{10}} = 46$ and ${a_5} + {a_7} = 52$

That gives us ${a_5} + {a_7} = a + 4d + a + 6d = 52$ i.e. $2a + 10d = 52$

Hence $a + 5d = 26$

Also we have ${a_{10}} = a + 9d = 46$

Therefore, $a + 9d - a - 5d = 46 - 26$

Hence, $4d = 20$ i.e. $d = 5$

That gives us, $a + 9\left( 5 \right) = 46$ i.e. $a = 1$

Therefore, the AP is $1,6,11,16,21,...$

8. Find the 20th term of an A.P. whose 7th term is 24 less than the 11th term, first term being 12.

Ans: Given that $a = 12$ and ${a_7} = {a_{11}} - 24$

That gives us,

$\Rightarrow {a_{11}} - {a_7} = a + 10d - a - 6d = 24$

Or,

$\Rightarrow 4d = 24$

Hence,

$\Rightarrow d = 6$

Now, the 20th term is given by

$\Rightarrow {a_n} = a + \left( {n - 1} \right)d$

i.e.

$\Rightarrow {a_{20}} = 12 + \left( {20 - 1} \right) \cdot \left( 6 \right)$

That is

$\Rightarrow {a_{20}} = 126$

9. If the 9th term of an A.P. is zero, prove that its 29th term is twice its 19th term.

Ans: Given that ${a_9} = 0$ i.e. $a + 8d = 0$ that gives us $a = - 8d$

Now we can write ${a_{29}} = a + 28d$ i.e. ${a_{29}} = - 8d + 28d$

Hence, ${a_{29}} = 20d$

Also, we have ${a_{19}} = a + 18d$ i.e. ${a_{19}} = - 8d + 18d$

Hence, ${a_{19}} = 10d$

Multiplying both sides by 2, we get

$\Rightarrow 2{a_{19}} = 20d$

That gives us,

$\Rightarrow {a_{29}} = 2{a_{19}}$

10. Find whether 55 is a term of the A.P.: 7, 10, 13, … or not. If yes, find which term it is.

Ans: Given terms are 7,10,13,...

Here $a = 7$ and $d = 3$

Now if ${a_n} = 55$ then we can write,\

$\Rightarrow 55 = 7 + \left( {n - 1} \right)3$

That gives,

$\Rightarrow \left( {n - 1} \right) = \dfrac{{48}}{3}$

i.e.

$\Rightarrow n = 17$

Hence 55 is the 17th term of the AP 7,10,13,...

11. Determine k so that $\left( {{k^2} + 4k + 8} \right)$  , $\left( {2{k^2} + 3k + 6} \right)$ and $\left( {3{k^2} + 4k + 4} \right)$ are three consecutive terms of an A.P.

Ans: The three given terms will be consecutive terms of an AP if the common difference among them is constant.

Here, we have

$\Rightarrow \left( {2{k^2} + 3k + 6} \right) - \left( {{k^2} + 4k + 8} \right) = {k^2} - k - 2$

And,

$\Rightarrow \left( {3{k^2} + 4k + 4} \right) - \left( {2{k^2} + 3k + 6} \right) = {k^2} + k - 2$

Now if the above two differences are equal then,

$\Rightarrow {k^2} - k - 2 = {k^2} + k - 2$

That gives us,

$\Rightarrow - k = k$

Hence,

$\Rightarrow k = 0$

12. Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.

Ans: Let the three consecutive terms of an AP be $a - d,a,a + d$

Now according to the question, their sum is 207.

Hence,

$\Rightarrow \left( {a - d} \right) + a + \left( {a + d} \right) = 207$

That gives us,

$\Rightarrow 3a = 207$

Hence,

$\Rightarrow a = 69$

Also, the sum of two smaller parts is 4623.

Hence,

$\Rightarrow \left( {a - d} \right)a = 4623$

That gives us,

$\Rightarrow \left( {69 - d} \right)\left( {69} \right) = 4623$

Or,

$\Rightarrow 69 - d = \dfrac{{4623}}{{69}}$

i.e

$\Rightarrow 69 - d = 67$

Hence,

$\Rightarrow d = 2$

Hence, the three numbers are $67,69,71$ .

13. The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.

Ans: The angles of a triangle are given to be in an A.P. Let these angles be $\left( {a - d} \right)^\circ ,a^\circ ,\left( {a + d} \right)^\circ$ respectively.

Now by the angle sum property of a triangle, we have

$\Rightarrow \left( {a - d} \right)^\circ + a^\circ + \left( {a + d} \right)^\circ = 180^\circ$

That gives us,

$\Rightarrow \left( {3a} \right)^\circ = 180^\circ$

i.e.

$\Rightarrow a^\circ = 60^\circ$

Also, the greatest angle is twice the least, so we have

$\Rightarrow \left( {a + d} \right)^\circ = 2\left( {a - d} \right)^\circ$

That gives us,

$\Rightarrow \left( {60 + d} \right)^\circ = 2\left( {60 - d} \right)^\circ$

Or,

$\Rightarrow 3d^\circ = \left( {120 - 60} \right)^\circ$

i.e.

$\Rightarrow d^\circ = 20^\circ$

Hence, the angles of the triangle are $40^\circ ,60^\circ ,80^\circ$ .

14. If nth terms of two A.P.s: 9, 7, 5, … and 24, 21, 18, … are same, then find the values of n. Also find that term.

Ans: Given that, two APs that are 9, 7, 5, … where $a = 9,{d_1} = - 2$

And 24, 21, 18, … where $b = 24,{d_2} = - 3$

According to the question, we have ${a_n} = {b_n}$

That gives us,

$\Rightarrow a + \left( {n - 1} \right){d_1} = b + \left( {n - 1} \right){d_2}$

Or,

$\Rightarrow 9 + \left( {n - 1} \right) \cdot \left( { - 2} \right) = 24 + \left( {n - 1} \right) \cdot \left( { - 3} \right)$

That is,

$\Rightarrow 11 - 2n = 27 - 3n$

Hence,

$\Rightarrow n = 16$

Now,

$\Rightarrow {a_{16}} = 9 + 15\left( { - 2} \right)$

Hence,

$\Rightarrow {a_{16}} = - 21$

Also, we have

$\Rightarrow {b_{16}} = 24 + 15\left( { - 3} \right)$

That gives us,

$\Rightarrow {b_{16}} = - 21$

15. If the sum of 3rd and the 8th terms of an A.P. is 7 and the sum of 7th and 14th terms is –3, find the 10th term.

Ans: Given that ${a_3} + {a_8} = 7$ i.e. $a + 2d + a + 7d = 7$ hence $2a + 9d = 7$

And, ${a_7} + {a_{14}} = - 3$ i.e. $a + 6d + a + 13d = - 3$ hence $2a + 19d = - 3$

Subtracting the above two obtained equations, we get

$\Rightarrow \left( {2a + 19d} \right) - \left( {2a + 9d} \right) = - 3 - 7$

That gives us,

$\Rightarrow 10d = - 10$

Hence,

$\Rightarrow d = - 1$

Now that gives, $2a + 9\left( { - 1} \right) = 7$ hence $a = 8$

Therefore, the 10th term of this AP is given by

$\Rightarrow {a_{10}} = 8 + \left( {10 - 1} \right)\left( { - 1} \right)$

Hence,

$\Rightarrow {a_{10}} = - 1$

16. Find the 12th term from the end of the A.P.: –2, –4, –6, …, –100.

Ans: Given AP is –2, –4, –6, …, –100.

Consider the reverse of this AP i.e. -100,...,-6,-4,-2.

Then we have $a = - 100$ and $d = 2$

Now we have to find the 12th term from the start of this AP.

Hence,

$\Rightarrow {a_{12}} = - 100 + \left( {12 - 1} \right)\left( 2 \right)$

That gives us,

$\Rightarrow {a_{16}} = - 100 + 22$

i.e.

$\Rightarrow {a_{16}} = - 78$

17. Which term of the A.P.: 53, 48, 43, … is the first negative term ?

Ans: Given AP is 53, 48, 43,… where $a = 53$ and $d = - 5$

We have to find its first negative term so let its nth term be negative i.e. ${a_n} < 0$ .

That gives us,

$\Rightarrow a + \left( {n - 1} \right)d < 0$

Or,

$\Rightarrow 53 + \left( {n - 1} \right)\left( { - 5} \right) < 0$

That is,

$\Rightarrow \left( {n - 1} \right) > \dfrac{{53}}{5}$

i.e.

$\Rightarrow n > \dfrac{{58}}{5}$

Or,

$\Rightarrow n > 11\dfrac{3}{5}$

Hence the smallest natural number such that $n > 11\dfrac{3}{5}$ is $n = 12$

So 12th term of this AP is the first negative term.

Also,

$\Rightarrow {a_{12}} = 53 + \left( {12 - 1} \right)\left( { - 5} \right)$

Hence,

$\Rightarrow {a_{12}} = - 2$

18. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ?

Ans: The first and last number between 10 and 300, which when divided by 4 leaves a remainder 3 are 11 and 299 respectively.

Hence, let $a = 11,d = 4$ and ${a_n} = 299$

Then we have

$\Rightarrow {a_n} = a + \left( {n - 1} \right)d$

Or,

$\Rightarrow 299 = 11 + \left( {n - 1} \right)4$

That is,

$\Rightarrow \left( {n - 1} \right) = \dfrac{{288}}{4}$

Hence,

$\Rightarrow n = 73$

19. Find the sum of the two middle most terms of an A.P. $\dfrac{{ - 4}}{3}, - 1,\dfrac{{ - 2}}{3},...4\dfrac{1}{3}$

Ans: Given AP is $\dfrac{{ - 4}}{3}, - 1,\dfrac{{ - 2}}{3},...4\dfrac{1}{3}$ where $a = - \dfrac{4}{3}$ and $d = - 1 + \dfrac{4}{3} = \dfrac{1}{3}$ and ${a_n} = 4\dfrac{1}{3}$

That gives us,

$\Rightarrow 4\dfrac{1}{3} = - \dfrac{4}{3} + \left( {n - 1} \right)\left( {\dfrac{1}{3}} \right)$

Or,

$\Rightarrow 17 = \left( {n - 1} \right)$

Hence,

$\Rightarrow n = 18$

Now the two middle most terms are $\dfrac{{18}}{2} = 9th$ and $\dfrac{{18}}{2} + 1 = 10th$ terms.

The sum of 9th and 10th is given by,

$\Rightarrow {a_9} + {a_{10}} = a + 8d + a + 9d$

That gives us,

$\Rightarrow {a_9} + {a_{10}} = 2a + 17d$

Or,

$\Rightarrow {a_9} + {a_{10}} = 2\left( { - \dfrac{4}{3}} \right) + 17\left( {\dfrac{1}{3}} \right)$

That is,

$\Rightarrow {a_9} + {a_{10}} = - \dfrac{8}{3} + \dfrac{{17}}{3}$

Hence,

$\Rightarrow {a_{16}} + {a_{17}} = \dfrac{9}{3} = 3$

20. The first term of an A.P. is –5 and last term is 45. If the sum of the terms of the A.P. is 120, then find the number of terms and the common difference.

Ans: Given that, $a = - 5$ , ${a_n} = 45$ and ${S_n} = 120$

Hence, we have

$\Rightarrow {S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$

That gives us

$\Rightarrow 120 = \dfrac{n}{2}\left( { - 5 + 45} \right)$

Or,

$\Rightarrow n = \dfrac{{240}}{{40}}$

Hence,

$\Rightarrow n = 6$

Now, since ${a_6} = 45$

Therefore,

$\Rightarrow {a_6} = a + 5d$

Or,

$\Rightarrow 45 = - 5 + 5d$

Hence,

$\Rightarrow d = 10$

21. Find the sum:

(i) 1 + (–2) + (–5) + (–8) + … + (–236)

Ans: Here we have $a = 1$ , $d = - 3$ and ${a_n} = - 236$

That gives us,

$\Rightarrow - 236 = 1 + \left( {n - 1} \right)\left( { - 3} \right)$

Hence,

$\Rightarrow n - 1 = \dfrac{{237}}{3}$

i.e.

$\Rightarrow n = 80$

Now,

$\Rightarrow {S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$

That gives,

$\Rightarrow {S_n} = 40\left( { - 235} \right)$

Hence,

$\Rightarrow {S_n} = - 9400$

(ii) $\left( {4 - \dfrac{1}{n}} \right) + \left( {4 - \dfrac{2}{n}} \right) + \left( {4 - \dfrac{3}{n}} \right) + ...$ upto n terms

Ans: Here, we have $\left( {4 - \dfrac{1}{n}} \right) + \left( {4 - \dfrac{2}{n}} \right) + \left( {4 - \dfrac{3}{n}} \right) + ...$ where $a = \left( {4 - \dfrac{1}{n}} \right).$ and $d = \dfrac{1}{n}.$

Therefore, the sum upto n terms is given by,

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

That gives us,

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2\left( {4 - \dfrac{1}{n}} \right). + \left( {n - 1} \right)\left( {\dfrac{1}{n}} \right)} \right]$

Or,

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {8 - \dfrac{2}{n}. - 1 + \dfrac{1}{n}} \right]$

That is,

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {7 - \dfrac{1}{n}.} \right]$

Hence,

$\Rightarrow {S_n} = \dfrac{{\left( {7n - 1} \right)}}{2}$

(iii) $\dfrac{{a - b}}{{a + b}} + \dfrac{{3a - 2b}}{{a + b}} + \dfrac{{5a - 3b}}{{a + b}} + ...$ upto 11 terms

Ans: Here we have $\dfrac{{a - b}}{{a + b}} + \dfrac{{3a - 2b}}{{a + b}} + \dfrac{{5a - 3b}}{{a + b}} + ...$ where ${a_1} = \dfrac{{a - b}}{{a + b}}$ and $d = \dfrac{{2a - b}}{{a + b}}$

Hence, the sum upto 11 terms is given by,

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

That gives us,

$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {2\left( {\dfrac{{a - b}}{{a + b}}} \right) + \left( {11 - 1} \right)\left( {\dfrac{{2a - b}}{{a + b}}} \right)} \right]$

Or,

$\Rightarrow {S_{11}} = \dfrac{{11}}{{2\left( {a + b} \right)}}\left[ {2a - 2b + 20a - 10b} \right]$

That is,

$\Rightarrow {S_{11}} = \dfrac{{11\left( {22a - 12b} \right)}}{{2\left( {a + b} \right)}}$

Hence,

$\Rightarrow {S_{11}} = \dfrac{{11\left( {11a - 6b} \right)}}{{\left( {a + b} \right)}}$

22. Which term of the A.P., –2, –7, –12, … will be –77 ? Find the sum of this A.P. upto the term –77.

Ans: Given AP is –2, –7, –12, … where $a = - 2$ , $d = - 5$ and ${a_n} = - 77$

Then, we have

$\Rightarrow - 77 = - 2 + \left( {n - 1} \right)\left( { - 5} \right)$

That gives us,

$\Rightarrow \left( {n - 1} \right) = \dfrac{{75}}{5}$

Hence,

$\Rightarrow n = 16$

Now,

$\Rightarrow {S_{16}} = \dfrac{{16}}{2}\left[ { - 2 - 77} \right]$

Hence,

$\Rightarrow {S_{16}} = 8\left( { - 79} \right)$

i.e.

$\Rightarrow {S_{16}} = - 632$

23. If ${a_n} = 3 - 4n$, then show that ${a_1},{a_2},{a_3},...$ from an A.P. Also find ${S_{20}}$ .

Ans: Given that ${a_n} = 3 - 4n$

Hence we have ${a_1} = 3 - 4 = - 1$ , ${a_2} = 3 - 8 = - 5$ , ${a_3} = 3 - 12 = - 9$ , ${a_4} = 3 - 16 = - 13$ and so on.

Therefore it gives us $a = - 1$ and $d = - 4$ and hence $- 1, - 5, - 9, - 13, - 17,...$ forms an AP.

Also we have

$\Rightarrow {S_{20}} = \dfrac{{20}}{2}\left[ {2\left( { - 1} \right) + \left( {20 - 1} \right)\left( { - 4} \right)} \right]$

That gives us,

$\Rightarrow {S_{20}} = 10\left[ { - 78} \right]$

Hence,

$\Rightarrow {S_{20}} = - 780$

24. In an A.P., if ${S_n} = n\left( {4n + 1} \right)$ then find the A.P.

Ans: Given that, in an AP we have ${S_n} = n\left( {4n + 1} \right)$

Hence, we have ${S_1} = 1 \cdot \left( {4\left( 1 \right) + 1} \right) = 5$ so we can write $a = 5$

Also ${S_2} = 2 \cdot \left( {4\left( 2 \right) + 1} \right) = 18$ so ${a_2} = 18 - 5 = 13$

Hence, we have $d = 13 - 5 = 8$

Therefore, the required AP is $5,13,21,29,37,45,...$

25. In an A.P. if ${S_n} = 3{n^2} + 5n$ and ${a_k} = 164$ , then find the value of k.

Ans: Given that, in an AP we have ${S_n} = 3{n^2} + 5n$

Hence, we have ${S_1} = 3{\left( 1 \right)^2} + 5\left( 1 \right) = 8$ so we can write $a = 8$

Also, ${S_2} = 3{\left( 2 \right)^2} + 5\left( 2 \right) = 22$ so we have ${a_2} = 22 - 8 = 14$

Hence, we have $d = 14 - 8 = 6$

Therefore, the required AP is $8,14,20,26,32,38,44,...$

Now if ${a_k} = 164$

Then we have

$\Rightarrow 164 = 8 + \left( {k - 1} \right)\left( 6 \right)$

That gives us,

$\Rightarrow \left( {k - 1} \right) = \dfrac{{156}}{6}$

Hence,

$\Rightarrow k = 27$

26. If ${S_n}$ denotes the sum of first n terms of an A.P., prove that ${S_{12}} = 3\left( {{S_8} - {S_4}} \right)$ .

Ans: We know that the sum of n terms of an AP is given by,

$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

That gives us,

$\Rightarrow {S_{12}} = \dfrac{{12}}{2}\left[ {2a + \left( {12 - 1} \right)d} \right]$

Hence,

$\Rightarrow {S_{12}} = 6\left[ {2a + 11d} \right]$

Also, we can write

$\Rightarrow {S_8} = \dfrac{8}{2}\left[ {2a + \left( {8 - 1} \right)d} \right]$

That is,

$\Rightarrow {S_8} = 4\left[ {2a + 7d} \right]$

And similarly,

$\Rightarrow {S_4} = \dfrac{4}{2}\left[ {2a + \left( {4 - 1} \right)d} \right]$

i.e.

$\Rightarrow {S_4} = 2\left[ {2a + 3d} \right]$

Now consider $3\left( {{S_8} - {S_4}} \right)$

That is,

$\Rightarrow 3\left( {{S_8} - {S_4}} \right) = 3\left[ {4\left( {2a + 7d} \right) - 2\left[ {2a + 3d} \right]} \right]$

That gives us,

$\Rightarrow 3\left( {{S_8} - {S_4}} \right) = 3\left[ {8a + 28d - 4a - 6d} \right]$

Or,

$\Rightarrow 3\left( {{S_8} - {S_4}} \right) = 3\left[ {4a + 22d} \right]$

i.e.

$\Rightarrow 3\left( {{S_8} - {S_4}} \right) = 6\left[ {2a + 11d} \right]$

Hence,

$\Rightarrow 3\left( {{S_8} - {S_4}} \right) = {S_{12}}$

27. Find the sum of first 17 terms of an A.P. whose 4th and 9th terms are –15, and –30 respectively.

Ans: Given that ${a_4} = - 15$ i.e. $a + 3d = - 15$

And ${a_9} = - 30$ i.e. $a + 8d = - 30$

Subtracting the above two equations gives us, $\left( {a + 8d} \right) - \left( {a + 3d} \right) = - 30 - \left( { - 15} \right)$ i.e. $d = - 3$

Hence $a + 3\left( { - 3} \right) = - 15$ i.e. $a + 3\left( { - 3} \right) = - 6$

Now the sum of first 17 terms is given by,

$\Rightarrow {S_{17}} = \dfrac{{17}}{2}\left[ {2\left( { - 6} \right) + \left( {17 - 1} \right)\left( { - 3} \right)} \right]$

That gives,

$\Rightarrow {S_{17}} = 17\left( { - 30} \right)$

Hence,

$\Rightarrow {S_{17}} = - 510$

28. If sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.

Ans: Given that, ${S_6} = 36$ i.e. $3\left( {2a + 5d} \right) = 36$ hence $2a + 5d = 12$

and ${S_{16}} = 256$ i.e. $8\left( {2a + 15d} \right) = 256$ hence $2a + 15d = 32$

Subtracting the above obtained two equations, we get

$\Rightarrow \left( {2a + 15d} \right) - \left( {2a + 5d} \right) = 32 - 12$

That gives us,

$\Rightarrow 10d = 20$

Hence,

$\Rightarrow d = 2$

That gives us, $2a + 5\left( 2 \right) = 12$ i.e. $a = 1$

Now the sum of the first 10 terms is given by,

$\Rightarrow {S_{10}} = \dfrac{{10}}{2}\left[ {2\left( 1 \right) + \left( {10 - 1} \right)\left( 2 \right)} \right]$

That gives us,

$\Rightarrow {S_{10}} = 5\left[ {2 + 18} \right]$

Hence,

$\Rightarrow {S_{10}} = 100$

29. Find the sum of all the 11 terms of an A.P. whose middle most term is 30.

Ans: Given that, the middle most term of an AP is 30 and the AP has 11 terms.

Hence, $\dfrac{{11 + 1}}{2} = \dfrac{{12}}{2} = 6$ i.e. the 6th term is 30 so ${a_6} = 30$

That gives us, $a + 5d = 30$

Now the sum of all the 11 terms of the AP is given by,

$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {2a + \left( {11 - 1} \right)d} \right]$

That gives us

$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {2a + 10d} \right]$

Or,

$\Rightarrow {S_{11}} = 11\left( {a + 5d} \right)$

That is,

$\Rightarrow {S_{11}} = 11\left( {30} \right)$

Hence,

$\Rightarrow {S_{11}} = 330$

30. Find the sum of last 10 terms of the A.P. 8, 10, 12, ..., 126.

Ans: Given AP is 8, 10, 12, ..., 126.

Consider the reverse of this AP i.e. 126,...12,10,8

Then we have $a = 126$ and $d = - 2$

Now the sum of first 10 terms of this AP is given by,

$\Rightarrow {S_{10}} = \dfrac{{10}}{2}\left[ {2\left( {126} \right) + \left( {10 - 1} \right)\left( { - 2} \right)} \right]$

That gives us,

$\Rightarrow {S_{10}} = 5\left[ {252 - 18} \right]$

Hence,

$\Rightarrow {S_{10}} = 1170$

31. Find the sum of first seven numbers which are multiples of 2 as well as of 9. [Hint: Take the L.C.M. of 2 and 9]

Ans: The first number which is a multiple of 2 and 9 is $2 \times 9 = 18$

Hence, the numbers which are multiples of 2 as well as of 9 are $18,36,54,72,90,...$

Here, we have $a = 18$ and $d = 18$

Now the sum of first seven numbers is given by,

$\Rightarrow {S_7} = \dfrac{7}{2}\left[ {2\left( {18} \right) + \left( {7 - 1} \right)\left( {18} \right)} \right]$

That gives us,

$\Rightarrow {S_7} = \dfrac{7}{2}\left[ {36 + 108} \right]$

i.e.

$\Rightarrow {S_7} = 7\left[ {72} \right]$

Hence,

$\Rightarrow {S_7} = 504$

32. How many terms of the A.P.: –15, –13, –11, … are needed to make the sum –55 ? Explain the reason for double answer.

Ans: Given AP is –15, –13, –11,… where $a = - 15$ , $d = 2$ and ${a_n} = - 55$

Therefore, we have

$\Rightarrow - 55 = \dfrac{n}{2}\left[ {2\left( { - 15} \right) + \left( {n - 1} \right)\left( 2 \right)} \right]$

That gives us

$\Rightarrow - 55 = n\left[ {n - 16} \right]$

Or,

$\Rightarrow {n^2} - 16n + 55 = 0$

That is,

$\Rightarrow {n^2} - 11n - 5n + 55 = 0$

i.e.

$\Rightarrow n\left( {n - 11} \right) - 5\left( {n - 11} \right) = 0$

That gives,

$\Rightarrow \left( {n - 11} \right)\left( {n - 5} \right) = 0$

Hence,

$\Rightarrow n = 11{\text{ or }}5$

Therefore, either 11 or 5 terms are needed to obtain the sum -55.

This is because, if we take $n = 11$ then we have

$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {2\left( { - 15} \right) + \left( {11 - 1} \right)\left( 2 \right)} \right]$

That gives us,

$\Rightarrow {S_{11}} = 11\left[ { - 15 + 10} \right]$

Hence,

$\Rightarrow {S_{11}} = - 55$

Also, if we take $n = 5$ then we have

$\Rightarrow {S_5} = \dfrac{5}{2}\left[ {2\left( { - 15} \right) + \left( {5 - 1} \right)\left( 2 \right)} \right]$

That gives,

$\Rightarrow {S_5} = 5\left[ { - 15 + 4} \right]$

Hence,

$\Rightarrow {S_5} = - 55$

33. The sum of first n terms of an A.P. whose first terms is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is –30, and the common difference is 8. Find n.

Ans: Given that, an AP where $a = 8$ and ${d_1} = 20$ and another AP where $a = - 30$ and ${d_2} = 8$

According to the question, we have

$\Rightarrow {S_n} = S{'_{2n}}$

That gives us,

$\Rightarrow \dfrac{n}{2}\left[ {2\left( 8 \right) + \left( {n - 1} \right)\left( {20} \right)} \right] = \dfrac{{2n}}{2}\left[ {2\left( { - 30} \right) + \left( {2n - 1} \right)\left( 8 \right)} \right]$

Or,

$\Rightarrow n\left[ {8 + \left( {n - 1} \right)\left( {10} \right)} \right] = n\left[ { - 60 + 16n - 8} \right]$

That is,

$\Rightarrow 10n - 2 = 16n - 68$

i.e.

$\Rightarrow 66 = 6n$

Hence,

$\Rightarrow n = 11$

34. Kanika was given her pocket money on Jan. 1, 2008. She puts ₹ 1 on day 1, ₹ 2 on day 2, ₹ 3 on day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent ₹ 204 of her pocket money, and found that at the end of the month she still had ₹ 100 with her. How much was her pocket money for the month ?

Ans: Let Kanika’s pocket money for the month be x rupees.

Now the money put by her in her piggy bank on each day in January is $1,2,3,4,5,....31$

Here, we have $a = 1$ , $d = 1$ and $n = 31$

Then the total amount in the piggy bank is given by,

$\Rightarrow {S_{31}} = \dfrac{{31}}{2}\left[ {2\left( 1 \right) + \left( {31 - 1} \right)\left( 1 \right)} \right]$

That gives us,

$\Rightarrow {S_{31}} = 31\left[ {1 + 15} \right]$

Hence,

$\Rightarrow {S_{31}} = 496$

Also, she had spent Rs.204 of her pocket money and in the end still has Rs.100.

Therefore, we have

$\Rightarrow x - 496 - 204 = 100$

That gives us,

$\Rightarrow x = 800$

Hence, Kanika’s monthly pocket money is Rs.800.

35. Yasmeen saves ₹ 32 during the first month, ₹ 36 in the second month and ₹ 40 in 3rd month. if she continues to save in this manner, in how many months will she save ₹ 2000 ?

Ans: The monthly savings of Yasmeen(in Rupees) is given as $32,36,40...$

Here, we have $a = 32$ , $d = 4$ and let ${a_n} = 2000$

Then we have,

$\Rightarrow 2000 = \dfrac{n}{2}\left[ {2\left( {32} \right) + \left( {n - 1} \right)\left( 4 \right)} \right]$

That gives us,

$\Rightarrow 2000 = n\left[ {32 + 2n - 2} \right]$

Or,

$\Rightarrow 2000 = n\left[ {30 + 2n} \right]$

That can be written as,

$\Rightarrow 2{n^2} + 30n - 2000 = 0$

Or,

$\Rightarrow {n^2} + 15n - 1000 = 0$

i.e.

$\Rightarrow {n^2} + 40n - 25n - 1000 = 0$

That is,

$\Rightarrow n\left( {n + 40} \right) - 25\left( {n + 40} \right) = 0$

That gives us,

$\Rightarrow \left( {n + 40} \right)\left( {n - 25} \right) = 0$

Hence,

$\Rightarrow n = - 40{\text{ or }}25$

Since n must be positive, therefore $n = 25$ .

Therefore, Yasmeen will save ₹ 2000 in 25 months.

Long Answer Questions Sample

Question 1: The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15. Find the numbers.

Ans: Given that, the sum of four consecutive numbers in an AP is 32.

Let these four consecutive numbers be $a - 3d,a - d,a + d,a + 3d$

Then we have,

$\Rightarrow a - 3d + a - d + a + d + a + 3d = 32$

That gives us

$\Rightarrow 4a = 32$

Or,

$\Rightarrow a = 8$

Now according to the question, we have

$\Rightarrow \dfrac{{\left( {a - 3d} \right)\left( {a + 3d} \right)}}{{\left( {a - d} \right)\left( {a + d} \right)}} = \dfrac{7}{{15}}$

That gives,

$\Rightarrow \dfrac{{{a^2} - {{\left( {3d} \right)}^2}}}{{{a^2} - {d^2}}} = \dfrac{7}{{15}}$

Or,

$\Rightarrow 15\left( {{a^2} - 9{d^2}} \right) = 7\left( {{a^2} - {d^2}} \right)$

That can be written as,

$\Rightarrow 15{a^2} - 7{a^2} = 135{d^2} - 7{d^2}$

Or,

$\Rightarrow 8{a^2} = 128{d^2}$

Now, since $a = 8$ , hence

$\Rightarrow 8 \times 64 = 128{d^2}$

i.e.

$\Rightarrow {d^2} = \dfrac{{8 \times 64}}{{128}}$

Or,

$\Rightarrow d = \sqrt 4$

Hence,

$\Rightarrow d = \pm 2$

Therefore, the required AP is either $2,6,10,14,...$ or $14,10,6,2,...$

Sample Question 2: Solve the equation :

1 + 4 + 7 + 10 +...+ x =287

Ans: Given that, $1 + 4 + 7 + 10 + ... + x = 287$

Here, we have $a = 1$ , $d = 3$ , ${a_n} = x$ and ${S_n} = 287$

So we can write it as,

$\Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$

That gives us,

$\Rightarrow 287 = \dfrac{n}{2}\left( {2\left( 1 \right) + \left( {n - 1} \right)\left( 3 \right)} \right)$

Or,

$\Rightarrow 574 = n\left( {2 + 3n - 3} \right)$

That gives us,

$\Rightarrow 3{n^2} - n - 574 = 0$

Or,

$\Rightarrow 3{n^2} + 41n - 42n - 574 = 0$

i.e.

$\Rightarrow n\left( {3n + 41} \right) - 14\left( {3n + 41} \right) = 0$

That gives,

$\Rightarrow \left( {3n + 41} \right)\left( {n - 14} \right) = 0$

Hence,

$\Rightarrow n = 14, - \dfrac{{41}}{3}$

Since n is a natural number, therefore we have $n = 14$

Now that gives us, $x = {a_{14}}$

Hence,

$\Rightarrow x = a + 13d$

i.e.

$\Rightarrow x = 1 + 13\left( 3 \right)$

Therefore,

$\Rightarrow x = 40$

EXERCISE 5.4

1. The sum of the first five terms of an A.P. and the sum of the first seven terms of the same A.P. is 167. If the sum of the first ten terms of this A.P. is 235, find the sum of its first twenty terms.

Ans: Given that ${S_5} + {S_7} = 167$ and ${S_{10}} = 235$

Now since ${S_5} + {S_7} = 167$ so we can write

$\Rightarrow \dfrac{5}{2}\left[ {2a + 4d} \right] + \dfrac{7}{2}\left[ {2a + 6d} \right] = 167$

That gives us,

$\Rightarrow 5\left[ {a + 2d} \right] + 7\left[ {a + 3d} \right] = 167$

Or,

$\Rightarrow 12a + 31d = 167$

Also since ${S_{10}} = 235$ , therefore we have

$\Rightarrow \dfrac{{10}}{2}\left[ {2a + 9d} \right] = 235$

That gives us,

$\Rightarrow 2a + 9d = 47$

Multiplying both sides by 6, we get

$\Rightarrow 12a + 54d = 282$

Now subtracting these two obtained equations, we get

$\Rightarrow \left( {12a + 54d} \right) - \left( {12a + 31d} \right) = 282 - 167$

That gives us,

$\Rightarrow 23d = 115$

Hence,

$\Rightarrow d = 5$

Now that gives us $12a + 31\left( 5 \right) = 167$ or $12a = 12$ or $a = 1$

Hence, the sum of its first twenty terms is given by,

$\Rightarrow {S_{20}} = \dfrac{{20}}{2}\left[ {2\left( 1 \right) + 19\left( 5 \right)} \right]$

That gives us,

$\Rightarrow {S_{20}} = 970$

2. Find the

(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.

Ans: The first and last integers between 1 and 500 which are multiples of 2 as well as of 5 are 10 and 490 respectively.

Here, $a = 10$ , $d = 10$ and ${a_n} = 490$

Now,

$\Rightarrow 490 = 10 + \left( {n - 1} \right)\left( {10} \right)$

That gives us,

$\Rightarrow n = 49$

Now,

$\Rightarrow {S_{49}} = \dfrac{{49}}{2}\left[ {2\left( {10} \right) + \left( {49 - 1} \right)\left( {10} \right)} \right]$

That gives us,

$\Rightarrow {S_{49}} = 49\left[ {10 + 48\left( 5 \right)} \right]$

Hence,

$\Rightarrow {S_{49}} = 49\left( {250} \right)$

i.e.

$\Rightarrow {S_{49}} = 12250$

(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.

Ans: The first and last integers from 1 and 500 which are multiples of 2 as well as of 5 are 10 and 500 respectively.

Here, $a = 10$ , $d = 10$ and ${a_n} = 500$

Now,

$\Rightarrow 500 = 10 + \left( {n - 1} \right)\left( {10} \right)$

That is,

$\Rightarrow n = 50$

Now,

$\Rightarrow {S_{50}} = \dfrac{{50}}{2}\left[ {10 + 500} \right]$

That is,

$\Rightarrow {S_{50}} = 25\left( {510} \right)$

Hence,

$\Rightarrow {S_{50}} = 12750$

(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.

[Hint (iii) : These numbers will be: multiples of 2+ multiples of 5 – multiples of 2 as well as of 5.]

Ans: Sum of those integers from 1 to 500 which are multiples of either 2 or 5 can be obtained by the sum of those integers from 1 to 500 which are multiples of 2 and sum of those integers from 1 to 500 which are multiples of 5 minus the sum of those integers from 1 to 500 which are multiples of both 2 and 5.

That is, the the required sum is $\left( {2 + 4 + 6 + ... + 500} \right) + \left( {5 + 10 + ... + 500} \right) - \left( {10 + 20 + ...500} \right)$

We can write is as $S = {S_1} + {S_2} - {S_3}$

Now for ${S_1}$ , we have $a = 2,d = 2,{a_n} = 500$

$\Rightarrow 500 = 2 + \left( {n - 1} \right)\left( 2 \right)$

That is,

$\Rightarrow n = 250$

Hence,

$\Rightarrow {S_1} = \dfrac{{250}}{2}\left[ {2 + 500} \right]$

That is,

$\Rightarrow {S_1} = 62750$

Now for ${S_2}$ , we have $a = 5,d = 5,{a_n} = 500$

$\Rightarrow 500 = 5 + \left( {n - 1} \right)\left( 5 \right)$

Hence,

$\Rightarrow n = 100$

Therefore,

$\Rightarrow {S_2} = \dfrac{{100}}{2}\left( {5 + 500} \right)$

i.e.

$\Rightarrow {S_2} = 25250$

Now for ${S_3}$ , we have $a = 10,d = 10,{a_n} = 500$

$\Rightarrow 500 = 10 + \left( {n - 1} \right)\left( {10} \right)$

Hence,

$\Rightarrow n = 50$

Therefore,

$\Rightarrow {S_3} = \dfrac{{50}}{2}\left( {10 + 500} \right)$

That is,

$\Rightarrow {S_3} = 12750$

Now, since $S = {S_1} + {S_2} - {S_3}$

Therefore,

$\Rightarrow S = 62750 + 25250 - 12750$

i.e.

$\Rightarrow S = 75250$

3. The 8th term of an A.P. is half its second term and 11th term exceeds one third of its fourth term by 1. Find the 15th term.

Ans: Given that $2{a_8} = {a_2}$ and ${a_{11}} = \dfrac{{{a_4}}}{3} + 1$

That gives us,

$\Rightarrow 2\left( {a + 7d} \right) = a + d$

i.e.

$\Rightarrow a + 13d = 0$

Or,

$\Rightarrow 2a + 26d = 0$

And, also we have

$\Rightarrow a + 10d = \dfrac{{\left( {a + 3d} \right)}}{3} + 1$

That gives,

$\Rightarrow 3a + 30d = a + 3d + 3$

i.e.

$\Rightarrow 2a + 27d - 3 = 0$

Subtracting these two equations, we get

$\Rightarrow \left( {2a + 27d - 3} \right) - \left( {2a + 26d} \right) = 0$

That gives us,

$\Rightarrow d = 3$

Therefore, $a + 13\left( 3 \right) = 0$ i.e. $a = - 39$

Now,

$\Rightarrow {a_{15}} = - 39 + \left( {15 - 1} \right)\left( 3 \right)$

That is,

$\Rightarrow {a_{15}} = 3$

4. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the A.P.

Ans: Given that $n = 37$ and the sum of the three middle most terms is 225 i.e. ${a_{18}} + {a_{19}} + {a_{20}} = 225$ and the sum of the last three is 429 i.e. ${a_{35}} + {a_{36}} + {a_{37}} = 429$

Now that gives us,

$\Rightarrow a + 17d + a + 18d + a + 19d = 225$

i.e.

$\Rightarrow 3a + 54d = 225$

That is,

$\Rightarrow a + 18d = 75$

And, also we have

$\Rightarrow a + 34d + a + 35d + a + 36d = 429$

That gives us,

$\Rightarrow 3a + 105d = 429$

i.e.

$\Rightarrow a + 35d = 143$

Now subtracting these two equations we get,

$\Rightarrow \left( {a + 35d} \right) - \left( {a + 18d} \right) = 143 - 75$

That gives us,

$\Rightarrow 17d = 68$

Hence,

$\Rightarrow d = 4$

Now that gives us, $a + 18\left( 4 \right) = 75$ i.e. $a = 3$

Therefore, the required AP is $3,7,11,15,19,...$

5. Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9.

[Hint (ii): These numbers will be: Total numbers – Total numbers divisible by 9.]

Ans: The integers between 100 and 200 that are divisible by 9 are $108,115,124,...198$

Here $a = 108$ , $d = 9$ and ${a_n} = 198$

Hence, we have

$\Rightarrow 198 = 108 + \left( {n - 1} \right)\left( 9 \right)$

That is,

$\Rightarrow n = 11$

Now we have,

$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {108 + 198} \right]$

That gives us,

$\Rightarrow {S_{11}} = 11\left[ {153} \right]$

i.e.

$\Rightarrow {S_{11}} = 1683$

(ii) First we have to find the sum of all the numbers between 100 and 200.

Hence we have $101,102,103,...199$ where $a = 101$ , $d = 1$ and ${a_n} = 199$

That gives us,

$\Rightarrow 199 = 101 + \left( {n - 1} \right)\left( 1 \right)$

Hence,

$\Rightarrow n = 99$

Now, we have

$\Rightarrow {S_{99}} = \dfrac{{99}}{2}\left[ {101 + 199} \right]$

That is,

$\Rightarrow {S_{99}} = 99\left[ {150} \right]$

i.e.

$\Rightarrow {S_{99}} = 14850$

Hence the sum of numbers between 100 and 200 which are not divisible by 9 is given by,

$\Rightarrow S = 14850 - 1683$

i.e.

$\Rightarrow S = 13167$

6. The ratio of the 11th term to the 18th term of an A.P. is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

Ans: Given that, ${a_{11}}:{a_{18}} = 2:3$

That gives us,

$\Rightarrow \dfrac{{a + 10d}}{{a + 17d}} = \dfrac{2}{3}$

Or,

$\Rightarrow 3\left( {a + 10d} \right) - 2\left( {a + 17d} \right) = 0$

Hence,

$\Rightarrow a - 4d = 0$

i.e.

$\Rightarrow a = 4d$

Now the ratio ${a_5}:{a_{21}}$ is given by,

$\Rightarrow \dfrac{{{a_5}}}{{{a_{21}}}} = \dfrac{{a + 4d}}{{a + 20d}}$

That gives,

$\Rightarrow \dfrac{{{a_5}}}{{{a_{21}}}} = \dfrac{{4d + 4d}}{{4d + 20d}}$

Hence,

$\Rightarrow \dfrac{{{a_5}}}{{{a_{21}}}} = \dfrac{1}{3}$

Also the ratio ${S_5}:{S_{21}}$ is given by,

$\Rightarrow \dfrac{{{S_5}}}{{{S_{21}}}} = \dfrac{{\dfrac{5}{2}\left[ {2a + \left( {5 - 1} \right)d} \right]}}{{\dfrac{{21}}{2}\left[ {2a + \left( {21 - 1} \right)d} \right]}}$

That gives us,

$\Rightarrow \dfrac{{{S_5}}}{{{S_{21}}}} = \dfrac{{5\left[ {a + 2d} \right]}}{{21\left[ {a + 10d} \right]}}$

Or,

$\Rightarrow \dfrac{{{S_5}}}{{{S_{21}}}} = \dfrac{{5\left[ {4d + 2d} \right]}}{{21\left[ {4d + 10d} \right]}}$

That is,

$\Rightarrow \dfrac{{{S_5}}}{{{S_{21}}}} = \dfrac{{5 \times 6d}}{{21 \times 14d}}$

Hence,

$\Rightarrow \dfrac{{{S_5}}}{{{S_{21}}}} = \dfrac{5}{{49}}$

7. Show that the sum of an A.P. whose first term is a, the second term b and the last term c, is equal to $\dfrac{{\left( {a + c} \right)\left( {b + c - 2a} \right)}}{{2\left( {b - a} \right)}}$ .

Ans: Given that, an AP which is $a,b,...c$

Here we have $a = a$ , $d = b - a$ and ${a_n} = c$

Now, we have

$\Rightarrow c = a + \left( {n - 1} \right)\left( {b - a} \right)$

That is,

$\Rightarrow n = \dfrac{{c - a + b - a}}{{b - a}}$

Hence,

$\Rightarrow n = \dfrac{{b + c - 2a}}{{b - c}}$

Now the sum of this AP is given by,

$\Rightarrow {S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$

That gives us,

$\Rightarrow {S_n} = \dfrac{{b + c - 2a}}{{2\left( {b - c} \right)}}\left( {a + c} \right)$

Hence,

$\Rightarrow {S_n} = \dfrac{{\left( {a + c} \right)\left( {b + c - 2a} \right)}}{{2\left( {b - c} \right)}}$

8. Solve the equation – 4 + (–1) + 2 + … + x = 437.

Ans: Given equation is, $- 4 + \left( { - 1} \right) + 2 + ... + x = 437$

Here, we have $a = - 4$ , $d = 3$ , ${a_n} = x$ and ${S_n} = 437$

Hence, we have

$\Rightarrow 437 = \dfrac{n}{2}\left[ {2\left( { - 4} \right) + \left( {n - 1} \right)\left( 3 \right)} \right]$

That gives,

$\Rightarrow 437 \times 2 = n\left[ { - 8 + 3n - 3} \right]$

Or,

$\Rightarrow 874 = 3{n^2} - 11n$

That is,

$\Rightarrow 3{n^2} - 11n - 874 = 0$

Or,

$\Rightarrow 3{n^2} + 46n - 57n - 874 = 0$

i.e.

$\Rightarrow n\left( {3n + 46} \right) - 19\left( {3n + 46} \right) = 0$

Hence,

$\Rightarrow \left( {3n + 46} \right)\left( {n - 19} \right) = 0$

That gives us,

$\Rightarrow n = - \dfrac{{46}}{3}{\text{ or }}19$

Therefore, we can take here the value $n = 19$ as it is a natural number.

Now, we have ${a_{19}} = x$

That gives us,

$\Rightarrow {a_{19}} = a + \left( {19 - 1} \right)d$

Or,

$\Rightarrow {a_{19}} = - 4 + \left( {18} \right)\left( 3 \right)$

Hence,

$\Rightarrow {a_{19}} = 50$

Therefore the value of x is $x = 50$ .

9. Jaspal Singh repays his total loan of ₹ 118000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month, what amount will be paid by him in the 30th instalment ? What amount of loan does he still have to pay after the 30th instalment ?

Ans: The monthly instalments paid by Jaspal Singh can be given as $1000,1100,1200,...$

Here we have $a = 1000$ , $d = 100$

Now his 30th instalment will be given by,

$\Rightarrow {a_{30}} = 1000 + 29\left( {100} \right)$

That gives us,

$\Rightarrow {a_{30}} = 3900$

Also, we have

$\Rightarrow {S_{30}} = \dfrac{{30}}{2}\left( {1000 + 3900} \right)$

i.e.

$\Rightarrow {S_{30}} = 15\left( {4900} \right)$

Hence,

$\Rightarrow {S_{30}} = 73500$

Now the amount of loan he still have to pay after the 30th instalment is,

$\Rightarrow S = 118000 - 73500$

That is,

$\Rightarrow S = 44500$

10. The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags.

Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ? What is the maximum distance she travelled carrying a flag ?

Ans: Given that, the number of flags is $n = 27$

Distance between two flags is given as 2m.

Flags are stored at the middle most position i.e. at $n = \dfrac{{27 + 1}}{2} = 14th$ position

Hence, 13 flags are to be fixed on either side of the 14th flag.

Now the distance covered by Ruchi in fixing the first flag and returning back to her original position  is  given by,

$\Rightarrow {S_1} = 2 + 2 = 4m$

Similarly,

$\Rightarrow {S_2} = 4 + 4 = 8m$

And,

$\Rightarrow {S_3} = 6 + 6 = 12m$

This pattern can be given by $4,8,12,16,20,...$ upto 13 terms

Here we have $a = 4$ , $d = 4$ and $n = 13$

Therefore, we can write

$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left[ {2\left( 4 \right) + \left( {13 - 1} \right)\left( 4 \right)} \right]$

That gives us,

$\Rightarrow {S_{13}} = 13\left[ {4 + 24} \right]$

Hence,

$\Rightarrow {S_{13}} = 364$

Now the total distance covered by Ruchi for fixing 27 flags will be given by,

$\Rightarrow 2{S_{13}} = 728m$

Also the maximum distance covered by Ruchi carrying a flag is given by ${a_{13}}$ where $a = 2,d = 2$

Hence,

$\Rightarrow {a_{13}} = 2 + 12\left( 2 \right)$

Therefore

$\Rightarrow {a_{13}} = 26m$

Difference Between the NCERT Exemplar and the NCERT Textbook

The Exemplar focuses on the clarity of your ideas. In all sections, you'll notice that the questions in the sample differ from those in the NCERT textual exercises and illustrations. This distinction will help you adjust to different styles of questions and familiarize yourself with the kind of twists to expect from your board paper. CBSE usually asks the exemplar the same questions. So, if you want to get a perfect mark on your math exam, the NCERT Exemplar is a must-have. Here are some highlights:

Information that is reliable- Experts write all NCERT Exemplar books after conducting a comprehensive study on the topics. The material contained therein is entirely genuine and will lead you in the proper route.

Basics that are both clear and strong- apart from the added knowledge of a higher level, NCERT Exemplar books cover the fundamentals of all areas for students. Whatever competitive exam you're preparing for, these books will help you understand the fundamentals and play a crucial role in your preparation.

FAQs on NCERT Exemplar for Class 10 Maths Chapter 5 - Arithmetic Progressions - Free PDF Download

1. What is Arithmetic Progression in Class 10 Math?

Arithmetic Progression (AP) is a numerical series in which the difference between any two subsequent numbers is a fixed value. For example, the natural number sequence 1, 2, 3, 4, 5, 6,... is an AP because the difference between two consecutive terms (say 1 and 2) is equal to one (2 -1). Even when dealing with odd and even numbers, the common difference between two consecutive words will be equal to 2.

2. What do sequences, series, and progression mean in NCERT Class 10 Math - Arithmetic Progression?

The meanings are as follows:

• A sequence is a set of integers that follow a defined pattern and might be finite or infinite. For example, the sequence 1, 2, 3, 4, 5,... is an infinite sequence of natural integers.

• The sum of the items in a sequence is called a series. The series of natural numbers 1+2+3+4+5... provides an example. A term is a number that appears in a sequence or series.

• A progression is a series of events in which the general term can be stated mathematically. You can visit Vedantu for more information.

3. What resources can I use to ace my Class 10 Math exam?

You can refer to previous years’ question papers, revision notes, practice exercises more from Vedantu. These notes and other resources are curated by our very experienced academics. These resources are very important for students aiming to score high in their exams.

4. Can I download the NCERT Exemplar Solutions for Class 10 Math - Arithmetic Progressions for free?

Yes, you can download NCERT Exemplar Solutions for Class 10 Math - Arithmetic Progressions absolutely free of cost. Vedantu doesn’t charge for these resources, they are available on the website free of cost. You can download the PDF any time you want.

5. Why should I download the NCERT Exemplar for Class 10 Math - Arithmetic Progressions?

You should download the NCERT Exemplar for Class 10 Math - Arithmetic Progressions because of the following reasons:

• All answers are written following CBSE norms.

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