$\text{ Let }{a}_1,a_2,a_3,\dots .,a_{49}\text{ be in A.P. such that }\sum _{k=0}^{12}{a}_{4k+1}=416\text{ and }{a}_9+a_{43}=66\text{ . If }$$a_1^2+a_2^2+\dots .+q_{17}^2=140m\text{ , then }m\text{ is equal to }$

[SolutionStep1]:

$\text{ Let }{a}_1,a_2,a_3,\dots \dots \dots ,a_{49}\text{ be in A.P }$

$\text{ Let ' }d\text{ ' be the common difference }$

[SolutionStep2]:

$\text{ Given }\sum _{k=0}^{12}{a}_{4k+1}=416$

$\Rightarrow a_1+a_5+a_9+\mathrm{....}+a_{49}=416$

$\text{ Here }n=13\text{ , common difference }D=4d$

[SolutionStep3]:

$\therefore S_n=\frac{n}{2}\left[2a+\left(n-1\right)D\right]$

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

$\Rightarrow a_1+24d=\mathrm{32.................}\left(1\right)$

$\text{ Given }{a}_9+a_{43}=66$

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

$\Rightarrow a_1+25d=\mathrm{33..............}\left(2\right)$

$\left(2\right)-\left(1\right)\Rightarrow d=1$

$\text{ From }\left(1\right),a_1=8$

[SolutionStep4]:

$\text{ Let }{a}_1^2+a_2^2+\dots .+a_{17}^2=S$

$\Rightarrow 8^2+9^2+\dots ..+(24{)}^2=S \left(\because a_{17}=a_1+16d=8+16\left(1\right)=24\right)$

$\Rightarrow S=1^2+2^2+\dots ..+7^2+8^2+9^2+\dots .+(24{)}^2-\left(1^2+2^2+\dots .+7^2\right)$

$=\frac{\left(24\right)\left(25\right)\left(49\right)}{6}-\frac{7\left(8\right)\left(15\right)}{6} \left(\because \sum n^2=\frac{n(n+1)(2n+1)}{6}\right)$

$=\frac{1}{6}\left[29400-840\right]$

$=4760$

$=\left(140\right)\left(34\right)$

$\therefore m=34$