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माना $a_{1}, a_{2}, a_{3}, \ldots ., a_{49}$ एक समांतर श्रेढ़ी में ऐसे है कि $\sum_{k=0}^{12} a_{4 k+1}=416$ तथा $a_{9}+a_{43}=66$ है। यदि $a_{1}^{2}+a_{2}^{2}+\ldots . .+a_{17}^{2}=140\, m$ है, तो $m$ बराबर है
$68$
$34$
$33$
$66$
Solution
(2) $\because$ $\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416 \Rightarrow \frac{{13}}{2}\left[ {2{a_1} + 48d} \right] = 416$
$ \Rightarrow {a_1} + 24d = 32\,\,\,\,\,\,\,\,……….\left( 1 \right)$
Now, ${a_9} + {a_{43}} = 66 \Rightarrow 2{a_1} + 50d = 66\,\,\,\,…….\left( 2 \right)$
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} \right)1} \right]}^2} = 140\,m} } $
$ \Rightarrow \sum\limits_{r = 1}^{17} {{{\left( {r + 7} \right)}^2} = 140\,m} $
$ \Rightarrow \sum\limits_{r = 1}^{17} {\left( {{r^2} + 14r + 49} \right)} = 140\,m$
$ \Rightarrow \left( {\frac{{17 \times 18 \times 35}}{6}} \right) + 14\left( {\frac{{17 \times 18}}{2}} \right) + \left( {49 \times 17} \right) = 140$
$ \Rightarrow m = 34$
