Question

The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1 : 5, find the AP. [CBSE 2014]

Answer 1

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

$$\begin{gathered} \therefore S_7=182 \\ \Rightarrow \frac{7}{2}\left(2a+6d\right)=182\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ \Rightarrow a+3d=26.....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} a_4:a_{17}=1:5\left(\mathrm{Given}\right) \\ \Rightarrow \frac{a+3d}{a+16d}=\frac{1}{5}\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 5a+15d=a+16d \end{gathered}$$
$\Rightarrow d=4a.....\left(2\right)$

Solving (1) and (2), we get

$$\begin{gathered} a+3\times 4a=26 \\ \Rightarrow 13a=26 \\ \Rightarrow a=2 \end{gathered}$$

Putting a = 2 in (2), we get

$d=4\times 2=8$

Hence, the required AP is 2, 10, 18, 26, ... .