Question:
If $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots \ldots . . \mathrm{a}_{\mathrm{n}}$ are in A.P. and $a_{1}+a_{4}+a_{7}+\ldots \ldots \ldots+a_{16}=114$, then $\mathrm{a}_{1}+\mathrm{a}_{6}+\mathrm{a}_{11}+\mathrm{a}_{16}$ is equal to :
Correct Option: , 3
Solution:
$a_{1}+a_{4}+a_{7}+a_{10}+a_{13}+a_{16}=114$
$\Rightarrow \frac{6}{2}\left(a_{1}+a_{16}\right)=114$
$\Rightarrow a_{1}+a_{16}=38$
So, $\mathrm{a}_{1}+\mathrm{a}_{6}+\mathrm{a}_{11}+\mathrm{a}_{16}=\frac{4}{2}\left(\mathrm{a}_{1}+\mathrm{a}_{16}\right)$
$=2 \times 38=76$
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