Let 1 2 3 49 a ,a ,a ,…,a be in A.P. such that 12 4k 1 k 0 a 416 + = å = and 9 43 a a 66. + = If 22 2 1 2 17 a a … a 140m + ++ = , then m is equal to :

Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{49}$ be in A.P. such that

$$

\begin{aligned}

&\sum_{\mathrm{k}=0}^{12} \mathrm{a}_{4 \mathrm{k}+\mathrm{l}}=416 \text { and } \mathrm{a}_{9}+\mathrm{a}_{43}=66 \text {. If } \\

&\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\ldots+\mathrm{a}_{17}^{2}=140 \mathrm{~m}, \text { then } \mathrm{m} \text { is equal to }

\end{aligned}

$$