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Question

${a_1} + {a_4} + {a_{  7}} + {a_{10}} + {a_{13}} + {a_{16}} = 114$

$ \Rightarrow \frac{6}{2}\left( {{a_1} + {a_{16}}} \right) = 114$

${a_1}

Vedclass · Accepted Answer

$76$

Vedclass · Answer

${a_1} + {a_4} + {a_{  7}} + {a_{10}} + {a_{13}} + {a_{16}} = 114$

$ \Rightarrow \frac{6}{2}\left( {{a_1} + {a_{16}}} ight) = 114$

${a_1} + {a_{16}} = 38$

So, ${a_1} + {a_6} + {a_{  11}} + {a_{16}} = \frac{4}{2}\left( {{a_1} + {a_{16}}} ight)$

$ = 2 	imes 38 \Rightarrow 76$

If $a_1, a_2, a_3 …………$ an 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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