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Question
Let ${a_1}$, ${a_2}$, ${a_3}$, ......... ,${a_{49}}$ be in A.P. such that

$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$ and ${a_9} + {a_{43}} = 66$.

$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$, then m is equal to
33
66
68
34

Solution

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