Let a1,a2,a3…an be in AP such that k=0∑12(a4k+1)=416 and a9+a43=66 If a12+a22+…+a172=140m then m is equal to

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Let {a}_{{1}},{a}_{{2}},{a}_{{3}}{a}_{{n}} be in AP such that

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Let $a_{1}, a_{2}, a_{3} \dots a_{n}$ be in AP such that $k = 0 \sum 12 (a_{4 k + 1}) = 416$ and $a_{9} + a_{43} = 66$ If $a_{1}^{2} + a_{2}^{2} + \dots + a_{17}^{2} = 140 m$ then m is equal to

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Sophia Discussed

Let

a_{1}, a_{2}, a_{3} \dots a_{n}

be in AP such that

k = 0 \sum 12 (a_{4 k + 1}) = 416

and

a_{9} + a_{43} = 66

a_{1}^{2} + a_{2}^{2} + \dots + a_{17}^{2} = 140 m

then m is equal to

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Question Text	Let $a_{1}, a_{2}, a_{3} \dots a_{n}$ be in AP such that $k = 0 \sum 12 (a_{4 k + 1}) = 416$ and $a_{9} + a_{43} = 66$ If $a_{1}^{2} + a_{2}^{2} + \dots + a_{17}^{2} = 140 m$ then m is equal to
Answer Type	Video solution: 1
Upvotes	151