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,.............