Question

In an A.P., if the 12^th term is −13 and the sum of its first four terms is 24, find the sum of its first ten terms ?

Answer 1

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

Given:

\[a_{12} = - 13\]
\[ S_4 = 24\]

\[a_{12} = - 13 \]
\[ \Rightarrow a + \left( 12 - 1 \right)d = - 13 \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow a + 11d = - 13 . . . . . (1)\]

Also,

\[S_4 = 24\]
\[ \Rightarrow \frac{4}{2}\left[ 2a + \left( 4 - 1 \right)d \right] = 24 \left[ S_n = \frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right] \right]\]
\[ \Rightarrow 2a + 3d = 12 . . . . . \left( 2 \right)\]

Solving (1) and (2), we get

a = 9 and d = −2

\[\therefore S_{10} = \frac{10}{2}\left[ 2 \times 9 + \left( 10 - 1 \right) \times \left( - 2 \right) \right]\]
\[ = 5\left( 18 - 18 \right)\]
\[ = 0\]

Hence, the sum of its first 10 terms is 0.