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.

Answer :

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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