Let $a_1, a_2, a_3,\dots \dots ,a_{49}$ be in $A.P.$ such that $\underset{k=0}{\overset{12}{\Sigma }}{a}_{4k+1}=416$ and $a_9+a_{43}=66.$ If $a_1^2+a_2^2+\dots +a_{17}^2=140m,$ then $m$ is equal to:

$a_1+a_5+ . . . +a_{49}=416$
Let common difference $= d$
$13a_1 +4d\left\{1+2+ . . . +12\right\}=416$

$13a_1 +4d\left(\frac{12\left(13\right)}{2}\right)=416$

$13a_1 +24d\left(13\right)=416$
$a_1+24d=32$      $. . . \left(i\right)$
$a_9+a_{43}=66$

$a_1+8d+a_1+42d=66$

$2a_1+50d=66$
$a_1+25d=33$      $. . . \left(ii\right)$
$a_1=8 ; d=1$
${\displaystyle \sum _{i=1}^{17}{a}_i{}^2}=140m$
$\Rightarrow a^2+{\left(a+d\right)}^2+ .. . +{\left(a+16d\right)}^2=140m$
$\Rightarrow 17a^2+d^2\left(1^2+2^2+ . . . +{16}^2\right)+2ad\left(1+2+ . . . + 16\right)=140m$

$\Rightarrow 17a^2+d^2\left(\frac{16\times 17\times 33}{6}\right)+2ad\left(\frac{16\times 17}{2}\right)=140m$
$\Rightarrow \left(17\times 64\right)+\left(8\times 11\times 17\right)+\left(8\times 16\times 17\right)=140m$
$\Rightarrow$ $m=34$