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If $a_{1}, a_{2}, a_{3}, - - - - - - - - - -, a_{n}$ are in A.P. and $a_{1} + a_{4} + a_{7} + - - - - - - + a_{16} = 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to

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Detailed Solution

$a_{1} + a_{4} + a_{7} + a_{10} + a_{13} + a_{16}$ also in A.P has sum $= \frac{6}{2} [a_{1} + a_{16}] = 114$
$a_{1} + a_{16} = 38$ . So $a_{1} + a_{6} + a_{11} + a_{16} = \frac{4}{2} (a_{1} + a_{16}) = 2 (38) = 76$

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