Question

The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference. [CBSE 2015]

Answer 1

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

$$\begin{gathered} a_4=11 \\ \Rightarrow a+\left(4-1\right)d=11\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+3d=11.....\left(1\right) \end{gathered}$$

Now,

$a_5+a_7=34$ (Given)
$$\begin{gathered} \Rightarrow \left(a+4d\right)+\left(a+6d\right)=34 \\ \Rightarrow 2a+10d=34 \\ \Rightarrow a+5d=17.....\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} 11-3d+5d=17 \\ \Rightarrow 2d=17-11=6 \\ \Rightarrow d=3 \end{gathered}$$

Hence, the common difference of the AP is 3.