Mathematics · 2019
JEE Main · 10 April 2019, Shift 1 · Q67
If a_1, a_2, a_3, … … …., a_n are in A.P. and a_1+a_4+a_7+… … ….+a_16=114, then a_1+ a_6+ a_11+ a_16 is equal to:
If $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots \ldots \ldots ., \mathrm{a}_{\mathrm{n}}$ are in A.P. and $\displaystyle a_1+a_4+a_7+\ldots \ldots \ldots .+a_{16}=114$, then $\displaystyle \mathrm{a}_1+\mathrm{a}_6+\mathrm{a}_{11}+\mathrm{a}_{16}$ is equal to :
Official answer
(3)
$\displaystyle 76$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.