Question

The 12^th 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, s₁₀

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

`= 5 xx (18-18) `

= 5 × 0

= 0

Hence, the required sum is 0.