Q. 31

# The sum of first 10 terms of an AP is – 150 and the sum of its next 10 terms is – 550. Find the AP.

Let a be the first term and d be the common difference.

Given: Sum of first 10 terms = S10 = - 150

Sum of next 10 terms = - 550

i.e. S20 - S10 = - 550

Consider S10 = - 150

(10/2)[2a + (10 - 1)d] = - 150

5 × [2a + 9d] = - 150

[2a + 9d] = - 30 ………………….(1)

Now, consider S20 - S10 = - 550

(20/2)[2a + (20 - 1)d] - (10/2)[2a + (10 - 1)d] = - 550

10 × [2a + 19d] - 5[2a + 9d] = - 550

10a + 145d = - 550 …………………..(2)

On subtracting equation (2) from 5 times of equation (2), we get,

- 100d = 400

d = - 4

a = 1/2 (-30 - 9d)

a = 1/2 (-30 + 36)

a = 3

Therefore the AP is 3, - 1, - 5, - 9,….

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