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. [CBSE 2015]

Answer 1

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

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

Also,

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

Solving (1) and (2), we get

$$\begin{gathered} 2\left(-13-11d\right)+3d=12 \\ \Rightarrow -26-22d+3d=12 \\ \Rightarrow -19d=12+26=38 \\ \Rightarrow d=-2 \end{gathered}$$

Putting d = −2 in (1), we get

$$\begin{gathered} a+11\times \left(-2\right)=-13 \\ \Rightarrow a=-13+22=9 \end{gathered}$$

∴ Sum of its first 10 terms, S₁₀

$$\begin{gathered} =\frac{10}{2}\left[2\times 9+\left(10-1\right)\times \left(-2\right)\right] \\ =5\times \left(18-18\right) \\ =5\times 0 \\ =0 \end{gathered}$$

Hence, the required sum is 0.