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: Sequence & Series (Mathematics)

\n\n\begin{aligned}\n& \text { (b) : Given, } a_1+a_4+a_7+a_{10}+a_{13}+a_{16}=114 \\\n& \Rightarrow\left(a_1+a_{16}\right)+\left(a_4+a_{13}\right)+\left(a_7+a_{10}\right)=114\n\n\end{aligned}\n\n

\n \n

\n\n\begin{gathered}\n\n\Rightarrow 3\left(a_1+a_{16}\right)=114\left[\because a_1+a_{16}=a_4+a_{13}=a_7+a_{10}\right] \\\n\n\Rightarrow a_1+a_{16}=\frac{114}{3}=38 \#(i)\n\n\end{gathered}\n\n

\n \nNow,

a_1+a_6+a_{11}+a_{16}=\left(a_1+a_{16}\right)+\left(a_6+a_{11}\right)

\n\n

\n\n\begin{aligned}\n& =2\left(a_1+a_{16}\right) \\\n& =2 \times 38 \\\n& =76 \\\n& \left(\because a_1+a_{16}=a_6+a_{11}\right)\n\n\end{aligned}\n\n

\n\n[Using (i)]\n

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