Let {a_1}, {a_2}, dots, {a_{49}} be in A.P. s | vedclass

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(B) Given $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$. This is a sum of $13$ terms in $A.P.$ with first term $a_1$ and common difference $4d$.$\frac{13}{2}

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$34$

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(B) Given $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$. This is a sum of $13$ terms in $A.P.$ with first term $a_1$ and common difference $4d$.$\frac{13}{2} [2a_1 + (13-1)4d] = 416$ $\Rightarrow \frac{13}{2} [2a_1 + 48d] = 416$ $\Rightarrow a_1 + 24d = 32 \dots (1)$Given ${a_9} + {a_{43}} = 66$ $\Rightarrow (a_1 + 8d) + (a_1 + 42d) = 66$ $\Rightarrow 2a_1 + 50d = 66$ $\Rightarrow a_1 + 25d = 33 \dots (2)$Subtracting $(1)$ from $(2)$,we get $d = 1$. Substituting $d=1$ in $(1)$,$a_1 + 24 = 32 \Rightarrow a_1 = 8$.Now,$\sum_{r = 1}^{17} a_r^2 = \sum_{r = 1}^{17} [8 + (r-1)1]^2 = \sum_{r = 1}^{17} (r+7)^2 = \sum_{r = 1}^{17} (r^2 + 14r + 49) = 140m$.Using summation formulas: $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$.For $n=17$: $\frac{17 	imes 18 	imes 35}{6} + 14 	imes \frac{17 	imes 18}{2} + 49 	imes 17 = 1785 + 2142 + 833 = 4760$.$140m = 4760 \Rightarrow m = \frac{4760}{140} = 34$.

Let ${a_1}, {a_2}, \dots, {a_{49}}$ be in $A.P.$ such that $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $\sum_{r = 1}^{17} a_r^2 = 140m$,then $m = \dots$

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