Let $a_1,a_2,a_3,\dots \dots \dots \dots \dots .$$a_{49}$be in A.P. such that$\sum _{k=0}^{12} a_{4k+1}=416\text{ and }{a}_9+a_{43}^{\mathrm{\prime }}=66 ,a_1^2+a_2^2+\dots \dots \dots ...+a_{17}^2=140m$ then $m$is equal to 

Solution:

$\sum _{k=0}^{12} a_{4k+1}=416$

$\Rightarrow a_1+a_5+\dots +a_{49}=416$ $\Rightarrow \frac{13}{2}\left(a_1+a_{49}\right)=416$

As $a_{49}=a_1+48d$ so we have

$\frac{13}{2}\left(2a_1+48d\right)=416\Rightarrow a_1+24d=32$     …………………... (i) 

Given, $a_9+a_{43}=66\Rightarrow a_1+8d+a_1+42d=66$

$\Rightarrow a_1+25d=33$         …………………………….…(ii)

The equation (i) and (ii) gives $a_1=8,d=1$

Now, $\begin{array}{l}{a}_1^2+a_2^2+\dots ..+a_{17}^2=8^2+9^2+\dots .+{24}^2\\ =\left(1^2+2^2+\dots ..+{24}^2\right)-\left(1^2+2^2+\dots .+7^2\right)\\ =\frac{24\cdot 25\cdot 49}{6}-\frac{7\cdot 8\cdot 15}{6}=4760=34\left(140\right)\end{array}$

$\therefore m=34$