If it is known that $a_{1} + a_{4} + a_{7} + a_{10} + a_{13} + a_{16} = 147$
then $a_{1} + a_{6} + a_{11} + a_{16} =$

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Solution

$\begin{matrix} a_{11} + a_{4} + a_{7} + a_{10} + a_{13} + a_{16} = 147 6 a + 45 r = 147 a_{1} + a_{6} + a_{11} + a_{16} = 4 a + 30 r = \frac{2}{3} (6 a + 45 r) \frac{2}{3} (147) = 98 \end{matrix}$

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Similar questions

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

a_{1}, a_{2}

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

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

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

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

26

169,

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

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