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$Let a_{1}, a_{2}, a_{3}, \dots a_{49} be in A.P such that \sum_{k = 0}^{12} a_{4 k + 1} = 416 and a_{9} + a_{43} = 66 . If$ ${a_{1}}^{2} + {a_{2}}^{2} + . . + {a_{17}}^{2} = 140 m then m is$

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Correct option is B

∑k=012 a4k+1=416⇒1322a1+48d=416⇒a1+24d=32 →1⇒a9+a43=66⇒2a1+50d=66 →21 and 2⇒a1=8,d1=1∑r=117 ar2=∑r=117 [8+(r−1)]2=140m⇒∑r=117 (r+7)2=140m⇒∑r=117 r2+49+14r=140m⇒17.18.356+14(17×18)2+49×17=140m⇒m=34

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Let a1,a2,a3,…a49 be in A.P such that ∑k=012 a4k+1=416 and a9+a43=66. If a12+a22+..+a172=140m then m is

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$Let a_{1}, a_{2}, a_{3}, \dots a_{49} be in A.P such that \sum_{k = 0}^{12} a_{4 k + 1} = 416 and a_{9} + a_{43} = 66 . If$ ${a_{1}}^{2} + {a_{2}}^{2} + . . + {a_{17}}^{2} = 140 m then m is$