Let $a_{1},\frac{a_{2}}{2},\frac{a_{3}}{2^{2}},\ldots,\frac{a_{10}}{2^{9}}$ be a G.P. of common rati

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

$\frac{a_{2}}{2a_{1}} = \frac{a_{3}}{2a_{2}} = \frac{a_{4}}{2a_{3}} = \ldots. = \frac{a_{10}}{2a_{9}} = \frac{1}{\sqrt{2}}$

$\therefore a_{1},a_{2},a_{3},,a_{10}$ are in G.P. with common ratio $\sqrt{2}$.

$${\sum_{i = 1}^{10}\mspace{2mu} a_{i} = \frac{a_{1}\left( (\sqrt{2})^{10} - 1 \right)}{\sqrt{2} - 1} = 62 }{\Rightarrow a_{1} = 2(\sqrt{2} - 1)}$$

Question

Options

Solution

More Progression (Sequence and Series) Questions