If a_1, a_2, a_3, dots, a_n are in A.P. and a | vedclass

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(A) The given sequence is an $A.P.$ with $6$ terms: $a_1, a_4, a_7, a_{10}, a_{13}, a_{16}$.The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(	ext{

Vedclass · Accepted Answer

$76$

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(A) The given sequence is an $A.P.$ with $6$ terms: $a_1, a_4, a_7, a_{10}, a_{13}, a_{16}$.The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(	ext{first term} + 	ext{last term})$.Here, $n = 6$, so $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = \frac{6}{2}(a_1 + a_{16}) = 114$.$3(a_1 + a_{16}) = 114 \Rightarrow a_1 + a_{16} = 38$.Now, we need to find the sum $a_1 + a_6 + a_{11} + a_{16}$.This is an $A.P.$ with $4$ terms, where the sum is $\frac{4}{2}(a_1 + a_{16})$.Substituting the value: $2 	imes 38 = 76$.

If $a_1, a_2, a_3, \dots, a_n$ are in $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 114$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

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