NCERT Solution Exercise 5.2 Class 10 Maths

Q1 Fill in the blanks in the following table, given that a is the first term d, the common difference and a_n the nth term of the AP:

(i)

a = 7

d = 3

n = 8

a_n = ?

Solution:

nth term of AP

a_n = a +  (n-1)d

a₈ = 7 + (8-1)×3

⇒ a₈ = 7 + 7×3

⇒ a₈ = 7 + 21

⇒ a₈ = 28

(ii)

a = -18

n = 10

a_n = 0

d = ?

Solution:

a_n = a + (n-1)d

⇒ 0 = -18 + (10-1)d

⇒ 18 = 9d

⇒ 18/9 = d

⇒ 2 = d

(iii) 

a =?

d = -3

n = 18

a_n = -5

Solution:

nth term of AP

a_n = a+(n-1)d

⇒ -5 = a + (18-1)×-3

⇒ -5 = a + 17 × -3

⇒ -5 = a -51

⇒ -5 + 51 = a

⇒ 46 = a

(iv)

a = -18.9

d = 2.5

a_n = 3.6 

n = ?

Solution:

nth term of AP

a_n = a + (n-1)d

⇒ 3.6 = -18.9 + (n-1)×2.5

⇒ 3.6 = -18.9 + 2.5n – 2.5

⇒ 3.6 = -21.4 + 2.5n

⇒ 3.6 + 21.4 = 2.5n

⇒ 25.0 = 2.5n

⇒ 25/2.5 = n

⇒ 10 = n

(v)

a = 3.5

d = 0

n = 105

a_n=?

Solution:

nth term of AP

a_n = a+(n-1)d

⇒ a_n = 3.5 +(105 -1)×0

⇒ a_n =3.5

Q.2 Choose the correct Choice in the following and justify:

(i) 30th term of the AP: 10,7,4,……is

(A) 97

(B) 77

(C) -77

(D) -87

Ans (C)

Explanation:

10, 7, 4, ……….

First term a = 10

d = Common difference  = 7-10 = -3

n = Number of terms  = 30

nth term of A.P.

a_n = a+(n-1)d

⇒ a₃₀ = 10 + (30-1)×-3

⇒ a₃₀ = 10 + 29×-3

⇒ a₃₀ = 10 – 87

⇒a_n = – 77

(ii) 11th term of the AP: -3, -1/2, 2, ………is

(A) 28

(B) 22

(C) -38

(D) -48½

Ans (B) 22

Explanation:

AP: -3, -1/2, 2, ……..

Number of terms = n = 11

a = 1st term = -3

d = Common difference = -1/2 -(-3) = -1/2 + 3 = (-1+6)/2 = 5/2

nth term of AP:

a_n = a+(n-1)d

⇒ a₁₁= -3+(11-1)×5/2

⇒ a₁₁= -3+(10)×5/2

⇒ a₁₁= -3+5×5

⇒ a₁₁= -3+25

⇒ a₁₁= 22

Q.3 In the following APs, find the missing terms in the boxes:-

(i) 2, ☐ , 26

Solution:

Middle term = (First term + 3rd term)/2

⇒ Middle term = 2+26/2 = 28/2 = 14

(ii) ☐ , 13, ☐, 3

Solution:

a = first term = ?

a₂ = 13

a₃ = ?

a₄ = 3 (4th term)

a + (n-1)d = a_n 

⇒ a + (2-1)d = a₂ 

⇒ a + d = 13 ———–(i)

a + (n-1)d = a₄

⇒ a + (4-1)d = 3

⇒ a + 3d = 3 ————(ii)

By Elimination Method

a + d = 13 ———–(i)

a + 3d = 3 ————(ii) Subtracting Equation(i) – Equation(ii)

We get

-2d = 10

⇒ d = 10/-2

⇒ d = -5 put into equation(i)

a + d = 13 ———–(i)

a -5 = 13

⇒ a = 13 + 5

⇒ a = 18

Third term = a+2d = 18 + 2×-5 = 18 -10 = 8

(iii) 5, ☐, ☐, 9½

Solution:

First term = a = 5

4th term = a₄ = 9½

⇒ a + 3d = 19/2

⇒ 5 + 3d = 19/2

⇒ 3d = 19/2 -5

⇒ 3d = (19-10)/2

⇒ 3d = 9/2

⇒ d = 9/6

⇒ d = 3/2

2nd term a₂ = a + d

⇒  a₂ = 5 + 3/2

⇒  a₂ = (10+3)/2

⇒ a₂ = 13/2

Third term a₃ = a+2d

⇒ a₃ = 5+2×3/2

⇒ a₃ = 5+3

⇒ a₃ = 8

(iv) -4, ☐, ☐, ☐, ☐, 6

First term = a = -4

sixth term = a₆ = 6

⇒ a+5d = 6 [ by using identity a_n = a+(n-1)d ]

⇒ -4+5d = 6

⇒ 5d = 6 + 4

⇒ 5d = 10

⇒ d = 10/5

⇒ d = 2

Now

Second term a₂ = a + d = -4+2 = -2

Third term = a₃ = a+2d = -4+2×2 = -4+4 = 0

Fourth Term = a₄ = a+3d = -4+3×2 = -4+6 = 2

5th term = a₄ +d = 2+2 = 4

Q.4 Which term of the AP: 3, 8, 13, 18, …………..is 78

Solution:

AP: 3, 8, 13, 18, ………………………………78

Let nth term of AP a_n= 78

First term = a = 3

Common difference = d = 8-3 = 5

nth term of AP:

a+(n-1)d = a_n

⇒ a+(n-1)d = 78

⇒ 3+(n-1)×5 = 78

⇒ (n-1)×5 = 78-3

⇒ (n-1) = 75/5

⇒ (n-1) = 15

⇒ n = 15+1

⇒ n = 16

Hence 16th term AP is 78.

Q5. Find the number of terms in each of the following APs:

(i) 7, 13, 19, ……………..205

Solution:

First term = a = 7

Common difference = d = 13-7=6

nth term a_n = 205

a+(n-1)d = a_n

⇒ 7+(n-1)×6 = 205

⇒ (n-1)×6 = 205 – 7

⇒ (n-1)×6 =198

⇒ (n-1) =198/6

⇒ (n-1) = 33

⇒ n = 33 +1

⇒ n = 34

Hence 34th term of given AP is 205

(ii) 18, 15½, 13, ………………-47

Solution:

First Term = a = 18

Common difference = 15½ – 18 = 31/2 – 18 = 31-36/2 =-5/2

nth term a_n = -47

a+(n-1)d = -47

⇒ 18+(n-1)×-5/2 = -47

⇒ (n-1)×-5/2 = -47-18

⇒ (n-1) = -65×2/-5

⇒ n-1 = 13×2

⇒ n-1 = 26

⇒ n = 26+1

⇒ n = 27

Hence 27th term of given AP is -47

Q.6 Check whether -150 is a term of the AP: 11, 8, 5, 2, ………………

Solution:

AP: 11, 8, 5, 2, …………………….-150

a = First term = 11

d = common difference = 8-11 = -3

Let nth term of AP = -150

a_n = -150

⇒ a+(n-1)d = -150

⇒ 11+(n-1)×-3 = -150

⇒ (n-1)×-3 = -150-11

⇒ (n-1)×-3 = -161

⇒ -3n+3 = -161

⇒ -3n = -161-3

⇒ -3n = -164

⇒ n = -164/-3

⇒ n = 52.6666

No -150 is not any term of given AP

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

Given:

11th term = 38

a₁₁ = 38

16th term = 73

a₁₆ =73

To Find:

31st term a₃₁ = ?

Solution:

a₁₁ = 38

⇒ a+(n-1)d = 38

⇒ a+(11-1)d = 38

⇒ a+10d = 38 ———-(i)

a₁₆ =73

⇒ a+(n-1)d = 73

⇒ a + (16-1) = 73

⇒ a + 15 = 73 —————-(ii)

By Elimination method

a+10d = 38 —————-(i)

a + 15d = 73 —————-(ii)

Subtracting Equation (ii) From Equation(i)

-5d = – 35

⇒ d = -35/-5

⇒ d = 7 put into equation (i)

a+10d = 38 —————-(i)

⇒ a + 10×7 = 38

⇒ a + 70 = 38

⇒ a = 38 – 70

⇒ a = – 32

31st term a₃₁ = a + 30d

⇒ a₃₁ = a +30d

⇒ a₃₁ = -32 + 30×7

⇒ a₃₁ = -32 + 210

⇒ a₃₁ = 178

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

Solution:

Number of terms = n = 50

3rd tern = 12

⇒ a + 2d = a₃

⇒ a + 2d = 12 ————————(i)

Last term = 106

a_n = 106

⇒ a₅₀ = 106

⇒ a + 49d = 106 ——————(ii)

By Elimination Method

a + 2d = 12 ———————(i)

a + 49d = 106 ——————(ii)

Subtracting (ii) from equation (i)

– 47d = -184

⇒ d = -94/-47

⇒ d = 2 Put into equation (i)

a + 2d = 12 —————-(i)

⇒ a + 2×2 = 12

⇒ a + 4 = 12

⇒ a = 12-4

⇒ a = 8

29th term = a₂₉ = a + 28d

⇒ a₂₉ = a + 28d

⇒ a₂₉ = 8 + 28×2

⇒ a₂₉ = 8 + 56

⇒ a₂₉ = 64

Hence 29th term of AP = 64

Q.9 If the 3rd and 9th terms of an AP are 4 and-8 respectively. Which term of this AP is Zero?

Solution:

3rd term = a₃ = 4

⇒ a + 2d = a₃

⇒ a + 2d = 4 ——————-(i)

9th term  = a₉ = -8

⇒ a + 8d = -8 ——————(ii)

By Elimination method

a + 2d = 4 ——————-(i)

a + 8d = -8 ——————(ii)

Subtracting (ii) From equation(i)

-6d = 12

⇒ d = 12/-6

⇒ d = -2 Put into equation (i)

a + 2d = 4 ——————-(i)

⇒ a + 2×-2 = 4

⇒ a – 4 = 4

⇒ a = 4 + 4

⇒ a = 8

Let nth term of AP is Zero

a_n = 0

a+(n-1)d = a_n 

⇒ a + (n-1)d = 0

⇒ 8 + (n-1)×-2 = 0

⇒ (n-1)×-2 = -8

⇒ n -1 = -8 /-2

⇒ n -1 = 4

⇒ n = 4 +1

⇒ n = 5

Hence 5th term is zero

Q. 10 The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

According to condition

17th term is exceed its 10th term by 7

a₁₇ = a₁₀ + 7

⇒ a + 16d = a + 9d + 7

⇒ a + 16d – a – 9d = 7

⇒ 7d = 7

⇒ d = 7/7

⇒ d = 1

Hence Common difference = 1

Q.11 Which term of the AP: 3, 15, 27, 39, ……….. will be 132 more than its 54th term?

Solution:

AP: 3, 25, 37, 39, ………..

First term = a = 3

Common difference = d = 15-3 = 12

Let nth term of AP will be 132 more than its 54th term

a_n = a₅₄ + 132

⇒ a + (n-1)d = a + 53d + 132

⇒ a + (n-1)d – a – 53d = 132

⇒ (n – 1 – 53)d = 132

⇒ (n – 54)×12 = 132

⇒ (n – 54) = 132/12

⇒ (n – 54) = 11

⇒ (n – 54 ) = 11

⇒ n = 11+54

⇒ n = 65

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

Solution:

AP Ist : a, a+d , a+2d, ….                                   AP IInd : A, A+d, A+2d, ……….

Common difference = d (Which is same in both APs)

Difference of their 100th term = 100

⇒ a₁₀₀ – A₁₀₀ = 100

⇒ a + 99d -A – 99d = 100

⇒ a – A = 100 ———(i)

a₁₀₀₀ = a + 999d

A₁₀₀₀ = A + 999d

a₁₀₀ – A₁₀₀ = a + 999d – ( A + 999d )

⇒ a₁₀₀ – A_n100 = a + 999d –  A – 999d

⇒ a₁₀₀ – A₁₀₀ = a –  A

⇒ a₁₀₀ – A₁₀₀ = a –  A

⇒ a₁₀₀ – A₁₀₀ = 100

Hence difference of 1000 terms also = 100

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

Solution:

AP: 105, 112, 119, …………………………994

a = 1st term = 105

d = Common difference = 112-105 = 7

a_n = nth term = 994

n = number of terms = ?

a_n = 994

⇒ a + (n-1)d = 994

⇒ 105 + (n-1)×7 = 994

⇒ (n-1)×7 = 994 – 105

⇒ (n-1)×7 = 889

⇒ (n-1) = 889/7

⇒ (n-1) = 127

⇒ n = 127+1

⇒ n = 128

Hence  there are 128 terms divisible by 7.

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

Solution:

AP: 12, 16, 20, …………….. , 248

a= first term = 12

d = common difference = 16-12 = 4

a_n = 248

a + (n-1)d = 248

⇒ 12 + (n-1)×4 = 248

⇒ (n-1)×4 = 248 – 12

⇒ (n-1)×4 = 236

⇒ (n-1) = 236/4

⇒ (n-1) = 59

⇒ n = 59 + 1

⇒ n = 60

Hence Total multiple of 4 lie between 10 and 250 = 60

Q.15 For what value of n, are the nth terms of two APs: 63, 65, 37, ……..and 3, 10, 17, ………….equal?

Solution:

API : 63, 65, 37, ……..

first term = a = 63

Common difference = d = 65-63 = 2

AP II: 3, 10, 17, …………

First term =a= 3

Common difference = d = 10 – 3 = 7

Let nth term are equal

a_nI = a_nII

a + (n-1)d = a + (n-1)d

⇒ 63 + (n-1)×2 = 3+ (n-1)×7

⇒ 63 + 2n – 2 = 3 + 7n – 7

⇒ 61 + 2n  = -4 + 7n

⇒ 2n -7n = – 4 – 61

⇒ -5n = – 65

⇒ n = -65/-5

⇒ n = 13

Hence 13th terms are equal.

Q.16 Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

3rd term = 16

a₃ = 16

⇒ a + 2d = 16 ———(i)

a₇ = a₅ + 12

⇒ a + 6d = a + 4d +12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12

⇒ d = 12/2

⇒ d = 6 Put into equation (i)

a + 2d = 16 ———(i)

⇒ a + 2×6 = 16

⇒ a + 12 = 16

⇒ a = 16 -12

⇒ a = 4

Hence AP: 4,10,16,22,…………

Q.17 Find the 2oth term from the last term of the AP: 3, 8, 13, ………,253.

Solution:

AP: 3, 8, 13, ………….,253

Last Term = l = 253

Common difference = d =8-3 = 5

n = 20

nth term of AP: (From last towards first)

l_n = l – (n-1)d

⇒ l₂₀ = 253 – (20-1)×5

⇒ l₂₀ = 253 – (19)×5

⇒ l₂₀ = 253 – 95

⇒ l₂₀ = 158

Hence 20th term From the last term is 158

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

Solution:

4th term + 8th Term = 24

a₄ + a₈ = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ 2(a + 5d) = 24

⇒ a + 5d = 24/2

⇒ a + 5d = 12 —————(i)

6th term + 10th term = 44

a₆ + a₁₀ = 44

⇒ a + 5d + a+ 9d = 44

⇒ 2a + 14d = 44

⇒ 2(a + 7d) = 44

⇒ a + 7d = 44/2

⇒ a + 7d = 22 ————–(ii)

By Elimination method

a + 5d = 12 —————(i)

a + 7d = 22 ————–(ii)

Subtract equation (ii) From equation (i)

5d – 7d = 12 – 22

⇒ -2d = -10

⇒ d = -10/-2

⇒ d = 5 Put into equation (i)

a + 5d = 12 —————(i)

⇒ a + 5×5 = 12

⇒ a + 25 = 12

⇒ a = 12 – 25

⇒ a = -13

Hence 1st three terms are

a1 = -13

a2 = -13 +5 =-8

a3 = -8 + 5 = -3

Q.19 Subba Rao started work in  1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7 000.

Given:

His starting salary = ₹ 5000

Increment each year = ₹ 200

To find:

In which year his salary = ₹ 7000 = ?

Solution:

First term = a = 5000

common difference d = 200

nth term of AP = a_n = 7000

Number if terms = n = ?

a_n = 7000

⇒ a + (n-1)d = 7000

⇒ 5000 + (n-1)×200 = 7000

⇒ (n-1)×200 = 7000 – 5000

⇒ n -1 = 2000/200

⇒ n = 10 + 1

⇒ n = 11

Hence in 11th year his Salary becomes = ₹ 7000

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

Given:

1st week saved  = ₹ 5

increased every week = ₹ 1.75

To Find:

In which week his savings becomes  = ₹ 20.75 = ?

Solution:

1st term = a = 5

Common difference = d = 1.75

nth term of AP = a_n = 20.75

Number of terms = n =?

a_n = 20.75

⇒ a +(n-1)d = 20.75

⇒ 5 + (n – 1)×1.75 = 20.75

⇒ (n-1)×1.75 = 20.75 – 5

⇒ (n -1) =15.75/1.75

⇒ (n-1) = 9

⇒ n = 9+1

⇒ n = 10

Hence in 10th week his weekly savings become = ₹ 20.75