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.

Let a be the first term and d be the common difference.

Given: a₁₂ = - 13

S₄ = 24

To find: Sum of first 10 terms.

Consider a₁₂ = - 13

⇒ a + 11d = - 13 ………………(1)

Also, S₄ = 24

⇒ (4/2) × [2a + (4 - 1)d] = 24

⇒ 2 × [2a + 3d] = 24

⇒ 2a + 3d = 12 …………….(2)

Subtracting equation (2) from twice of equation (1), we get,

19d = - 38

⇒ d = - 2

Now, from equation (1), we get

a = - 13 - 11d

⇒ a = - 13 - 11(-2)

⇒ a = - 13 + 22

⇒ a = 9

Now, Sum of first n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of first 10 terms of this arithmetic series is given by:

∴ S₁₀ = [2(9) + (10 - 1)(-2)]

= 5 × [18 - 18]

= 0

∴ S₁₀ = 0