Question

The 9th term of an AP is −32 and the sum of its 11th and 13th terms is −94. Find the common difference of the AP. [CBSE 2015]

Answer 1

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

$$\begin{gathered} a_9=-32 \\ \Rightarrow a+\left(9-1\right)d=-32\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+8d=-32.....\left(1\right) \end{gathered}$$

Now,

$a_{11}+a_{13}=-94$ (Given)
$$\begin{gathered} \Rightarrow \left(a+10d\right)+\left(a+12d\right)=-94 \\ \Rightarrow 2a+22d=-94 \\ \Rightarrow a+11d=-47.....\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} -32-8d+11d=-47 \\ \Rightarrow 3d=-47+32=-15 \\ \Rightarrow d=-5 \end{gathered}$$

Hence, the common difference of the AP is −5.