Question

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

Answer 1

Let a be the first term and d be the common difference of the AP. Then,

s₁₀ = -150                                 (Given)

`⇒ 10/2 (2a + 9d) = -150            { s_n = n/2 [ 2a + (n-1) d]}`

⇒ 5 (2a + 9d)= -150

⇒ 2a + 9d = -30              ................(1)

It is given that the sum of its next 10 terms is - 550.

Now,

s₂₀ Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms =

-150+(-550)= -700

∴ s₂₀ = -700

`⇒  20/2 (2a +19d )= -700`

⇒ 10(2a + 19d ) = -700

⇒ 2a +19d = -70          ...................(2)

Subtracting (1) from (2), we get

(2a + 19d ) - (2a + 9d) = -70 - (-30)

⇒ 10d = -40

⇒ d =-4

Putting d   = - 4 in (1), we get

2a + 9 × (-4) = -30

⇒ 2a = -30 + 36 = 6

⇒ a =3

Hence, the required AP is 3,-1,-5,-9,.............