Mathematics · 2020
JEE Main · 6 September 2020, Shift 2 · Q56
The common difference of the A.P. b_1, b_2, …, b_m is 2 more than the common difference of A.P. a_1, a_2, …, a_n. If a_40=-159, a_100=-399 and b_100=…
The common difference of the A.P. $\displaystyle \mathrm{b}_1, \mathrm{~b}_2, \ldots, \mathrm{~b}_{\mathrm{m}}$ is $\displaystyle 2$ more than the common difference of A.P. $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_{\mathrm{n}}$. If $\displaystyle \mathrm{a}_{40}=-159$, $\displaystyle \mathrm{a}_{100}=-399$ and $\displaystyle \mathrm{b}_{100}=\mathrm{a}_{70}$, then $\displaystyle \mathrm{b}_1$ is equal to :
Official answer
(2)
$\displaystyle -81$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.