Find $a_{1} + a_{6} + a_{11} + a_{16}$ if it is known that $a_{1}, a_{2} . . . .$ is an A.P. and $a_{1} + a_{4} + a_{7} + . . . + a_{16} = 147.$

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Solution

We have $a_{1}, a_{2}, . . .$ in A.P. with common difference d.
So, $a_{1} + a_{4} + a_{7} + . . . a_{16} = 147$
$\Rightarrow a_{1} + (a_{1} + 3 d) + . . . + (a_{1} + 15 d) = 147$
$\Rightarrow 6 a_{1} + 3 d (1 + 2 + . . . + 5) = 147$
$\Rightarrow 6 a_{1} + 45 d = 147$
$\Rightarrow 2 a_{1} + 15 d = 49$
Now, $a_{1} + a_{6} + a_{11} + a_{16} = 4 a_{1} + 30 d = 2 (2 a_{1} + 15 d) = 98$

Suggest Corrections

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}

a_{1}, a_{2}, a_{3}

a_{4} + a_{8} + a_{12} + a_{16} = 224

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