Let a_1, frac{a_2}{2}, frac{a_3}{2^2}, ldots, | vedclass

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(A) The given sequence is a $G$.$P$. with common ratio $\frac{1}{\sqrt{2}}$.Thus, $\frac{a_2/2}{a_1} = \frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}}$.Thi

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$2(\sqrt{2}-1)$

Vedclass · Answer

(A) The given sequence is a $G$.$P$. with common ratio $\frac{1}{\sqrt{2}}$.Thus, $\frac{a_2/2}{a_1} = \frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}}$.This implies $\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \frac{1}{\sqrt{2}}$, so $\frac{a_2}{a_1} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Therefore, $a_1, a_2, \ldots, a_{10}$ is a $G$.$P$. with first term $a_1$ and common ratio $R = \sqrt{2}$.The sum of the first $10$ terms is $S_{10} = a_1 \frac{R^{10}-1}{R-1} = 62$.$a_1 \frac{(\sqrt{2})^{10}-1}{\sqrt{2}-1} = 62$.$a_1 \frac{2^5-1}{\sqrt{2}-1} = 62$.$a_1 \frac{31}{\sqrt{2}-1} = 62$.$a_1 = \frac{62(\sqrt{2}-1)}{31} = 2(\sqrt{2}-1)$.

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

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