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JEE MAIN 2020
06-09-2020 S2
Question
The common difference of the A.P. $b_1, b_2, \ldots, b_m$ is 2 more than the common difference of A.P. $\mathrm{a}_1, \mathrm{a}_2$, $\_\_\_\_$ an. If $\mathrm{a}_{40}=-159, \mathrm{a}_{100}=-399$ and $\mathrm{b}_{100}=\mathrm{a}_{70}$, then $\mathrm{b}_1$ is equal to :
Select the correct option:
A
–127
B
–81
C
127
D
81
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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