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

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Question

(2)  $\because$ $\sum\limits_{k = 0}^{12} {{a_{4k + 1}}}  = 416 \Rightarrow \frac{{13}}{2}\left[ {2{a_1} + 48d} \right] = 416$

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Vedclass · Accepted Answer

$34$

Vedclass · Answer

(2)  $\because$ $\sum\limits_{k = 0}^{12} {{a_{4k + 1}}}  = 416 \Rightarrow \frac{{13}}{2}\left[ {2{a_1} + 48d} ight] = 416$

$ \Rightarrow {a_1} + 24d = 32\,\,\,\,\,\,\,\,..........\left( 1 ight)$

Now,  ${a_9} + {a_{43}} = 66 \Rightarrow 2{a_1} + 50d = 66\,\,\,\,.......\left( 2 ight)$

form eq. $(1)$ & $(2)$ we get; $d=1$ and ${a_1} = 8$

Also, $\sum\limits_{r = 1}^{17} {a_r^2 = \sum\limits_{r = 1}^{17} {{{\left[ {8 + \left( {r - 1} ight)1} ight]}^2} = 140\,m} } $

$ \Rightarrow \sum\limits_{r = 1}^{17} {{{\left( {r + 7} ight)}^2} = 140\,m} $

$ \Rightarrow \sum\limits_{r = 1}^{17} {\left( {{r^2} + 14r + 49} ight)}  = 140\,m$

$ \Rightarrow \left( {\frac{{17 	imes 18 	imes 35}}{6}} ight) + 14\left( {\frac{{17 	imes 18}}{2}} ight) + \left( {49 	imes 17} ight) = 140$

$ \Rightarrow m = 34$

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

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