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

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(A) The given sum is $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114$.This is an arithmetic progression with $6$ terms,where the first term is $a_1$

Vedclass · Accepted Answer

$76$

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(A) The given sum is $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114$.This is an arithmetic progression with $6$ terms,where the first term is $a_1$ and the last term is $a_{16}$.The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(first + last)$.Thus,$\frac{6}{2}(a_1 + a_{16}) = 114$.$3(a_1 + a_{16}) = 114 \Rightarrow a_1 + a_{16} = 38$.We need to find the sum $S = a_1 + a_6 + a_{11} + a_{16}$.This is also an arithmetic progression with $4$ terms,where the first term is $a_1$ and the last term is $a_{16}$.$S = \frac{4}{2}(a_1 + a_{16}) = 2(38) = 76$.

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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