Let a_1, (a_2)/(2), (a_3)/(2^2), , a_102^9 be a G.P.

Question

Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \ldots + a_{10} = 62$, then $a_1$ is equal to:

Accepted Answer

$\begin{gathered} \frac{{a}_{2}}{2{a}_{1}}=\frac{{a}_{3}}{2{a}_{2}}=\frac{{a}_{4}}{2{a}_{3}}=\ldots =\frac{{a}_{10}}{2{a}_{9}}=\frac{1}{\sqrt{2}} \ 	herefore {a}_{1},{a}_{2},{a}_{2},\ldots .{a}_{10}\sqrt{2} \ 	herefore \sum _{i=1}^{10}{a}_{i}=\frac{{a}_{1}((\sqrt{2}{)}^{10}-1)}{\sqrt{2}-1}=62\Rightarrow {a}_{1}=2(\sqrt{2}-1) \end{gathered}$

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