# Ncert Solutions For Class 10 Maths Ex 5.2

## Ncert Solutions For Class 10 Maths Chapter 5 Ex 5.2

Q.1 In the given A.P: 26, 40, 54, 68 ………………. which term is 208?

Sol.

From the given A.P:

a = 26,

d = 40 – 26 = 14,

an  = 208,

n = ??

Since,             an = a + (n – 1)d

Therefore, 208 = 26 + (n – 1)×14

14n = 208 + 14 – 26

Hence,            n = 14

14th term of this given A.P is 208

Q2. Find how many total numbers of terms are there in each of the following A.P

(i) 17, 26, 35, 44,……………………….179

Sol.

From the given A.P :

a = 17,

d  = (26 – 17) = 9,

an = 179

n = ??

since,         an = a + (n – 1)d

Therefore, 179 = 17 + (n – 1)×9

9n = 179 – 17 + 9

Therefore n = 19

Hence, there are 19 terms in this given A.P

(ii)13,223,433,.3373$\frac{1}{3},\frac{22}{3},\frac{43}{3},…………………….\frac{337}{3}$

From the given A.P :

a = 13$\frac{1}{3}$,

d = 22313=7$\frac{22}{3} – \frac{1}{3} = 7$,

an = 3373$\frac{337}{3}$,

n = ??

Since, an = a + (n – 1)d

Therefore, 3373=13+(n1)7$\frac{337}{3}=\frac{1}{3}+(n-1)7$

21n=337+21-1

Therefore n= 17

Hence, there are 17 terms in this given A.P

Q3. Seventh term of an A.P is -1 and fourth term is 41. Determine the A.P and hence find its 17th term.

Sol.

Given, a7 = -1 and a4 = 41

Let a be the first term and d be the common difference of an A.P

-1 = a + (7 – 1)d

-1 = a + 6d . . . . . . .  . . (1)

And   41 = a + (4 – 1)d

41 = a + 3d . . . . . . . (2)

Now, equation (1) – equation (2)

-1 – 41 = 6d – 3d    Therefore,   d = -14

Putting d = -14 in equation (2) we get

41 = a + 3×(-14)

Therefore, a = 83

Therefore the required A.P is 83, 69, 55, 12, 41, 27, 13………

Now, from the given A.P:

a = 83,

d = 69 – 83 = -14

n = 17

a17 = ??

Since,                   an = a + (n – 1)d

Therefore          a17 = 83 + (17 – 1)× -14

a17 = 208 + 14 – 26

a17 = -141

Therefore 17th term of this given A.P is -141

Q.4 There are 26 terms in an A.P of which fifth term is 61 and the last term is 292. Find 12th term

Sol.

Given,

a5 = a + (5-1)d = 61

61 = a + 4d…………………………………(1)

a26 = a + (26 – 1)d = 292

292 = a + 25d……………………………(2)

Equation (2) – Equation (1)

292 – 61 = 25d – 4d

Therefore, d = 11

Putting the value of d in equation (1) we get

61 = a + 4 × 11

Therefore, a = 17

Hence a12 = 17 + (12 – 1)×11 = 138

Therefore 12th term of this given A.P is 138

Q.5 Find which term of an A.P is zero, if fourth term and tenth of an A.P are -26 and 52 respectively.

Sol.

Given,

an = 0

a4 = a + (4 – 1)d = -26

-26 = a + 3d………………………………(1)

a10 = a + (10 – 1)d = 52

52 = a + 9d……………………………(2)

Equation (2) – Equation (1) we get:

52 + 26 = 9d – 3d

Therefore, d = 13

On putting the values of d in equation (1) we get:

-26 = a + 3×13

Therefore, a = -65

Now, let nth term be 0

Therefore,  0 = a + (n – 1)d

0 = -65 + (n – 1)×13

0 = -65 + 13n -13

n = 6

Therefore 6th term of the given A.P is zero.

Q.6 Find the common difference d, if 14th term of an A.P exceeds its 7th term by 49.

Sol.

Since,             an = a + (n – 1)d

Therefore,   a14 = a + (14 – 1)d

Similarly,      a7 = a + (7 – 1)d

Now, according to the given condition:

a14 – a7  = 49

49 = {a + (14 – 1)d} – {a + (7 – 1)d}

49 = 13d – 6d

Therefore, d = 7 (Common Difference)

Q.7 Consider a series : 3, 20, 37, 54,71 …………………………..,Which term of this given A.P will be 289 more than its 32nd term.

Sol.

Given, a = 3

d = 20 – 3 = 17

Therefore,  a32 = a + (32 – 1)×d

a32 = 3 + 31×17

a32 = 530

Now, according to the given condition:

an – a32 = 289    (Since, an > a32)

(3+ (n – 1)17) – 530 = 289

819 = 3 + 17n – 17

n = 49

Thus 49th term of this given A.P will be 289 more than its 32nd term.

Q.8 Find the first three terms of an A.P, if sum of 12th and 15th term is 219 and sum of 5th and 7th term is 114.

Sol.

Since, an = a + (n-1)d

Therefore, according to the given conditions:

a12+ a15 = 219

[a + (12-1)d] + [a + (15-1)d] = 219

2a + 25d = 219 ……………………………………………. (1)

And, a5 + a7 = 114

[a +(5-1)d] + [a+(7-1)d] = 114

2a + 10d = 114 ……………………………………………………(2)

Subtracting equation (2) from equation (1) we get

25d – 10d = 105

Therefore, d = 7

Putting values of d in equation (2) we get

2a + 10×7 = 114

Therefore a = 22

Hence 1st term of this A.P = a = 22

2nd term of this A.P = 22+(2-1)×7 = 29

3rd term of this A.P = 22+(3-1)×7 = 36

Q.9 Nikhil started his work in 2002 at an annual salary of Rs 9000 and each year he received an increment of Rs 300. Find in which year his income will reach Rs 12900?

Sol.

Since, each year Nikhil’s salary is increased by Rs 300 and his starting salary was Rs 9000.

Therefore, this forms an A.P with a = 9000 and d = 300

Given, an = 12900 and n = ??

Since, an = a +(n-1)d

Therefore, 12900 = 9000 + (n-1)×300

300n = 12900 + 300 – 9000

Therefore n = 14

Hence, the salary of Nikhil will be 12900 Rs after 14 years from 2002.

Q.10 Find the first three terms of an A.P, if sum of 11th and 14th term is 126 and sum of 4th and 6th term is 66.

Sol.

Since, an = a + (n-1)d

Therefore, according to the given conditions:

a11+ a14 = 126

[a + (11-1)d] + [a + (14-1)d] = 126

2a + 23d = 126 ……………………………………………. (1)

And, a4 + a6 = 66

[a +(4-1)d] + [a+(6-1)d] = 66

2a + 8d = 66 ……………………………………………………(2)

Subtracting equation (2) from equation (1) we get

23d – 8d = 60

Therefore, d = 4

Putting values of d in equation (2) we get

2a + 8×4 = 66

Therefore a = 17

Hence 1st term of this A.P = a = 17

2nd term of this A.P = 17+(2-1)×4 = 21

3rd term of this A.P = 17+(3-1)×4 = 25

Q.11 In the first week of a particular year Deepak saved Rs 10 and then he increased his weekly savings by 2.25Rs. If Deepak was able to increase his weekly savings to Rs 46 in nth week. Find n

Sol.

Money saved by Deepak in 1st week = 10Rs

Money saved by Deepak in 2nd week = 10 + 2.25 = 12.25Rs

Money saved by Deepak in 3rd week = 12.25 + 2.25 =14.50Rs

Therefore, the above given situation forms an A.P: 10, 12.25, 14.50, 16.75…………..

Here, a = 10

d = 2.25

an = 46

Since,       an= a+(n-1)d

Therefore 46= 10+(n-1)2.25

2.25n=46-10+2.25

Therefore n= 17

Hence in 17th week if that year Deepak was able to increase his weekly savings to Rs 46.

Q.12 Find 26th term from the last term of A.P: 7, 12, 17,22………..187.

Sol.

Here, a = 187 (In this case, last term will be first term)

d= -5 (since the given A.P is reversed, therefore d = 12 -17 = 7 – 12 = -5)

And, n = 26

Since, an= a + (n – 1)d

Therefore an = 187 + (26 – 1)×-5

an= 62

Therefore 26th term from the last term is 62.

Q.13 Consider two A.Ps :- 9, 15, 21, 27,…….. and 63,66, 69, 72,…….. Find the value of n, such that the nth term of both the A.P’s is equal.

Sol.

For First A.P, a = 9 and d = 15-9 = 6

Since          an = a+(n-1)d

Therefore an= 9+(n-1)6

(Or)             an= 6n+3

For second A.P, a = 63 and d = 66-63 = 3

Since,          an = a + (n – 1)d

Therefore,  an = 63 + (n – 1)×3

(Or)              an = 3n +60

Now, according to the given condition:

3n+60=6n+3

Therefore n = 19

Therefore, 19th term of both the A.P’s is equal.

Q.14 Find how many three digit numbers are divisible by 9

Sol.

First 3- digit number divisible by 9 = 108

Second 3- digit number divisible by 9 = 117

Similarly last 3-digit number divisible by 9 = 999

Hence it forms an A.P with d = 9,

108, 117, 126……………999

Now, from the above A.P:

a= 108, d= 9, an= 999

n=??

Since, an= a+ (n-1)×d

Therefore 999= 108+ (n-1)×9

9n=999-108+9

Therefore n= 900

Hence, there are 900 three digit numbers that are divisible by 9.

Formula for Sum of first n terms:

Let us suppose that there are n terms in an A.P, therefore sum of first n terms of an A.P is given by Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Here, a = first term, d = common difference and n = number of terms whose sum is to be determined.

REMEMBER:     an = Sn – Sn–1

Example-1: In an A.P -2, 5, 12, 19,……… Find the sum of first 8 terms.

Sol.

Here, a = -2 and   d = (12-5) = 7 and n = 8

Since,        Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore Sn = 82$\frac{8}{2}$ [2×-2+(8-1)7]

= 4 × [45]

Therefore sum of first 8 terms = 180.

Example – 2 How many terms of an A.P 34, 43, 52, 61,………………… must be taken so that there sum is equal to 1002.

Sol.

Given,  a = 34, d = 9 and Sn = 1002

Since, Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore, Sn = 1002 = n2$\frac{n}{2}$ [2×34+(n-1)×9]

1002×2 = n×[59 + 9n]

9n2 + 59n – 2004 = 0

Now, from the above quadratic equation:    a = 9, b = 59 and c = -2004

Substituting the values of a, b and c in quadratic formulae we get:

n=(59)+(59)24(9×2004)2×9$n=\frac{-(59)+\sqrt{(59)^{2}-4(9\times-2004)}}{2\times 9}$  and  n= (59)(59)24(9×2004)2×9$\frac{-(59)-\sqrt{(59)^{2}-4(9\times-2004)}}{2\times 9}$

n=59+3481+7214418$n=\frac{-59+\sqrt{3481+72144}}{18}$  and  n=593481+7214418$n=\frac{-59-\sqrt{3481+72144}}{18}$

n=59+7562518$n=\frac{-59+\sqrt{75625}}{18}$  and  n=597562518$n= \frac{-59-\sqrt{75625}}{18}$

n = 59+27518$\frac{-59+275}{18}$  and  n= 5902518$\frac{-590-25}{18}$

n=12     (neglecting the negative terms)

Therefore, the sum of first 12 terms in the given A.P will be 2004.

Example-3 Find the sum of first 100 odd positive integers.

Sol.

Odd positive integers can be written as: 3, 5, 7, 9, 11……

Here a = 3, d = 2 and n = 100

Since          Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore, Sn = 1002$\frac{100}{2}$ [2×3+(100-1)×2]

= 50[6+198]

= 10200

Therefore the sum of first 100 odd positive integers = 10200

Example – 4 The nth term of any particular series in given by (3n – 5). Find sum of first 16 terms.

Sol.

Given,             an = 3n – 5

Therefore,      a1 = 3×1 – 5 = -2

a2 = 3×2 – 5 = 1

a3 = 3×3 – 5 = 4

Therefore, the series becomes -2, 1, 4 . . . . .

Since   [1 – (-2)] = (4 – 1) = 3 = d

Hence, the above series forms an A.P with a = -2 and d = 3

Since, the sum of first n terms is given by: Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Sn = 162$\frac{16}{2}$ [2×-2+(16-1)×3]

= 8(-4 + 45)

= 328

Therefore the sum of first 16 terms is 328