$\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 }$${{\mathrm{a}}_1}^2+{{\mathrm{a}}_2}^2+..+{{\mathrm{a}}_{17}}^2=140\mathrm{m}\text{ then }\mathrm{m}\text{ is }$

$\sum _{\mathrm{k}=0}^{12} {\mathrm{a}}_{4k+1}=416\Rightarrow \frac{13}{2}\left[2{\mathrm{a}}_1+48\mathrm{d}\right]=416$

$\begin{array}{l}\Rightarrow {\mathrm{a}}_1+24\mathrm{d}=32 \to \left(1\right)\\ \Rightarrow {\mathrm{a}}_9+{\mathrm{a}}_{43}=66\Rightarrow 2{\mathrm{a}}_1+50\mathrm{d}=66 \to \left(2\right)\\ \left(1\right)\text{ and }\left(2\right)\Rightarrow {\mathrm{a}}_1=8,{\mathrm{d}}_1=1\\ \sum _{\mathrm{r}=1}^{17} {a_r}^2=\sum _{\mathrm{r}=1}^{17} [8+(\mathrm{r}-1){]}^2=140\mathrm{m}\\ \Rightarrow \sum _{\mathrm{r}=1}^{17} (\mathrm{r}+7{)}^2=140\mathrm{m}\\ \Rightarrow \sum _{\mathrm{r}=1}^{17} \left({\mathrm{r}}^2+49+14\mathrm{r}\right)=140\mathrm{m}\\ \Rightarrow \frac{\mathrm{17.18.35}}{6}+\frac{14(17\times 18)}{2}+49\times 17=140\mathrm{m}\\ \Rightarrow \mathrm{m}=34\end{array}$