# 10th CBSE math solution for exercise 5.3 part 3

This page 10th CBSE math solution for exercise 5.3 part 3 is going to provide you solution for every problems that you find in the exercise no 5.3 part 3

## 10th CBSE math solution for exercise 5.3 part 3 (ii) Given a = 7, a₁₃ = 35 find d and S₁₃

Solution:

a n = a + (n - 1) d

a₁₃ = 7 + (13 - 1) d

35 - 7 = 12 d

28/12 = d

d = 7/3

S n = (n/2) [a + l]

S₁₃ = (13/2) [7 + 35]

= (13/2) (42)

= 13 (21)

= 273

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

Solution:

a n = a + (n - 1) d

a₁₂ = a + (12 - 1) 3

37 =  a + 11 (3)

37 = a + 33

37 - 33 = a

a = 4

S n = (n/2) [a + l]

S₁₂ = (12/2) [4 + 37]

= 6 (41)

= 246

In the page 10th CBSE math solution for exercise 5.3 part 3 next we are going to see the solution of next problem.

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

Solution:

a n = a + (n - 1) d

a = a + (3 - 1) d

15 =  a + 2 d

a + 2 d = 15 -----(1)

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

S₁₀ = (10/2) [ 2 a + (10-1) d]

125 = 5 [2a + 9 d]

125/5 = 2 a + 9 d

2 a + 9 d = 25  -----(2)

(1) x 2 => 2 a + 4 d = 30

2 a + 9 d = 25

(-)     (-)     (-)

----------------

- 5 d = 5

d = - 1

Substitute d = -1 in the first equation

a + 2 (-1) = 15

a = 15 + 2

a = 17

a₁₀ = a + 9 d

= 17 + 9 (-1)

= 17 - 9

= 8

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

Solution:

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

S₉ = (9/2) [ 2a + (9 - 1) 5]

75 = (9/2) [ 2 a + 8 (5) ]

75 x (2/9) = 2 a + 40

(50/3) -40 = 2 a

2 a = (50 - 120)/3

2 a = -70/3

a = - 35/3

a= a + 8 d

= (-35/3) + 8 (5)

= (-35/3) +  40

= (-35 + 120)/3

= 85/3

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