Let a_1, a_2, , a_10 be positive real numbers such that [(a_2)/(2a_1) = (a_3)/(2a_2) = = a…

Question

Let $a_1, a_2, \ldots, a_{10}$ be positive real numbers such that
$$\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \cdots = \frac{a_{10}}{2a_9} = \frac{1}{\sqrt2},$$
and $a_1+a_2+\cdots+a_{10} = 62$. Then $a_1$ is equal to:

Accepted Answer

The given relations give $\dfrac{a_{i+1}}{a_i} = \sqrt2$ for each $i$, so $a_1,\dots,a_{10}$ is a G.P. with common ratio $\sqrt2$.

$$\sum_{i=1}^{10} a_i = \frac{a_1\big((\sqrt2)^{10}-1\big)}{\sqrt2-1} = 62 \implies a_1\cdot\frac{31}{\sqrt2-1} = 62 \implies a_1 = 2(\sqrt2-1).$$

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