Offline paper
Mathematics · 2018
JEE Main 2018 · 2018-04-08 Offline · Q68
Let a_1, a_2, a_3, ….., a_49 be in A.P. such that Σ_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 equal to:
Let $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots . ., \mathrm{a}_{49}$ be in A.P. such that $\displaystyle \sum_{\mathrm{k}=0}^{12} \mathrm{a}_{4 \mathrm{k}+1}=416$ and $\displaystyle \mathrm{a}_9+\mathrm{a}_{43}=66$. If $\displaystyle \mathrm{a}_1^2+\mathrm{a}_2^2+\ldots . .+\mathrm{a}_{17}^2=140 \mathrm{~m}$, then m is equal to :
Official answer
(3)
$\displaystyle 34$
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JEE Main 2018 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.