Question

In an A.P., $S_5+S_7=167$ and $S_{10}=235$,then find the A.P., where $S_n$ denotes the sum of its first n terms.

Answer 1

Given, $S_5+S_7=167$

$\Rightarrow \frac{5}{2}(2a+4d)+\frac{7}{2}(2a+6d)=167$

$\Rightarrow \frac{5}{2}\times 2(a+2d)+\frac{7}{2}\times 2(a+3d)=167$

$\Rightarrow 5a+10d+7a+21d=167$

$\Rightarrow 12a+31d=167...\left(i\right)$

$\Rightarrow \frac{10}{2}(2a+9d)=235$

$\Rightarrow 10a+45d=235$

$\Rightarrow 2a+9d=47...\left(ii\right)$

On multiplying equation (ii) by 6, we get:

$12a+54d=282$ ...(iii)

On subtracting equation (i) from (ii), we get:

$\begin{array}{cc}& \\ & 12a+54d=282\\ & 12a+31d=167\\ & ---\\ & \underset{–}{}\\ & 23d=115\\ \Rightarrow & d=5\end{array}$

Substituting value of d in equation (i), we get

$12a+31\times 5=167$

$12a+155=167$

$\Rightarrow 12a=12$

$\Rightarrow a=1$

Hence A.P. is 1,6,11......