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Sum of n Terms

If the sum of...

Question

If the sum of first $26$ terms of an A.P. is $169,$ then the value of $a_{1} + a_{6} + a_{11} + a_{16} + a_{21} + a_{26}$ is

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Solution

We know that,
$S = \frac{n}{2} (a_{1} + a_{26})$
$\Rightarrow 169 = \frac{26}{2} \times k$
$\Rightarrow k = 13$

$a_{1} + a_{6} + a_{11} + a_{16} + a_{21} + a_{26} = (a_{1} + a_{26}) + (a_{6} + a_{21}) + (a_{11} + a_{16}) = 3 k = 39$

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26

169,

a_{1} + a_{6} + a_{11} + a_{16} + a_{21} + a_{26}

a_{n}

a_{1} + a_{4} + a_{7} + . . . + a_{16} = 147

a_{1} + a_{6} + a_{11} + a_{16} =

a_{1} + a_{6} + a_{11} + a_{16}

a_{1}, a_{2} . . . .

a_{1} + a_{4} + a_{7} + . . . + a_{16} = 147.

a_{1} + a_{6} + a_{11} + a_{16}

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a_{1} + a_{4} + a_{7} + a_{10} + a_{13} + a_{16} = 147

a_{1}, a_{2}, a_{3}, \dots, a_{n}

a_{1} + a_{4} + a_{7} + \dots + a_{16} = 114

a_{1} + a_{6} + a_{11} + a_{16}

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