This page 10th CBSE math solution for exercise 5.3 part 7 is going to provide you solution for every problems that you find in the exercise no 5.3 part 7

10th CBSE math solution for exercise 5.3 part 7

(9) If the sum of 7 terms of an AP is 49 and that of 17 terms is 289,find the sum of first n terms.

Solution:

S₇= 49

S₁₇= 289

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

S₇= (7/2) [2 a + (7 - 1) d]

49 = (7/2) [ 2a + 6d]

(49 x 2)/7 = 2 a + 6 d

14 = 2 a + 6 d

a + 3 d = 7 -----(1)

S₁₇= (17/2) [2 a + (17 - 1) d]

289 = (17/2) [ 2a + 16d]

(289 x 2)/17 = 2 a + 16 d

34 = 2 a + 16 d

a + 8 d = 17 -----(1)

(1) - (2)

a + 3 d = 7

a + 8 d = 17

(-) (-) (-)

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- 5 d = -10

d = 2

Substitute d = 2 in the first equation

a + 3 (2) = 7

a + 6 = 7

a = 7 - 6

a = 1

To find the sum of first n terms, we have to apply the values of a and in the Sn formula

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

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

=(n/2) [2 + 2 n - 2]

=(n/2) [2 n]

= n²

(10) Show that a₁,a₂,............ an form an AP where an is defined as below

(i) a n = 3 + 4 n

(ii) a n = 9 - 5 n

Also find the sum of 15 terms in each case.

Solution:

(i) a n = 3 + 4 n

n = 1

a₁ = 3 + 4(1)

= 7

n = 2

a₂ = 3 + 4(2)

= 11

d = a₂ - a₁

= 11 - 7

= 4

So the series will be in the form 7 + 11 + ...........