Question

The 12th term of an AP is –13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.

Answer 1

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

a₁₂ = –13 

⇒ a + 11d = –13   ...(1) [a_n = a + (n – 1)d]

Also, 

S₄ = 24

⇒ `4/2 (2a + 3d) = 24`   ...`{S_n = n/2 [ 2a + (n - 1)d]}`

⇒ 2a + 3d = 12   ...(2)

Solving (1) and (2), we get

2(–13 – 11d) + 3d = 12

⇒ –26 – 22d + 3d = 12

⇒ –19d = 12 + 26 = 38 

⇒ d = –2 

Putting d = –2 in (1), we get

a + 11 × (–2) = –13

⇒ a = –13 + 22 = 9

∴ Sum of its first 10 terms,

`S_10 = 10/2 [2 xx 9 + (10 - 1) xx (-2)]`

= 5 × (18 – 18)

= 5 × 0

= 0

Hence, the required sum is 0.