Question ID: #693
Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \dots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_{1}+a_{2}+\dots+a_{10}=62$, then $a_{1}$ is equal to:
- (1) $2(\sqrt{2}-1)$
- (2) $2-\sqrt{2}$
- (3) $\sqrt{2}-1$
- (4) $2(2-\sqrt{2})$
Solution:
The common ratio of the given G.P. is obtained by dividing consecutive terms:
$$\frac{\frac{a_2}{2}}{a_1} = \frac{\frac{a_3}{2^2}}{\frac{a_2}{2}} = \dots = \frac{1}{\sqrt{2}}$$
$$\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \dots = \frac{1}{\sqrt{2}}$$
$$\frac{a_2}{a_1} = \frac{2}{\sqrt{2}} = \sqrt{2}$$
$$\frac{a_3}{a_2} = \frac{2}{\sqrt{2}} = \sqrt{2}$$
This shows that $a_1, a_2, a_3, \dots, a_{10}$ are in a G.P. with a common ratio $R = \sqrt{2}$.
Sum of $n$ terms of a G.P. is $S_n = \frac{a(R^n – 1)}{R – 1}$.
$$\sum_{i=1}^{10}a_{i} = \frac{a_1((\sqrt{2})^{10} – 1)}{\sqrt{2} – 1} = 62$$
$$62 = \frac{a_1(2^5 – 1)}{\sqrt{2} – 1}$$
$$62 = \frac{a_1(32 – 1)}{\sqrt{2} – 1}$$
$$62 = \frac{31a_1}{\sqrt{2} – 1}$$
$$2 = \frac{a_1}{\sqrt{2} – 1}$$
$$a_1 = 2(\sqrt{2} – 1)$$
Ans. (1)
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