Let a 1 , a 2 2 , a 3 2 2 , … , a 10 2 9 be a G.P. of common ratio 1 2 . If a 1 + a 2 + ? + a 10 = 62 , then a 1 is equal to: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(2)

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

$∴$ $a_{1}$

$\sum_{i = 1}^{10} a_{i} = \frac{a_{1} ({(\sqrt{2})}^{10} - 1)}{\sqrt{2} - 1} = 62$

$\Rightarrow a_{1} = 2 (\sqrt{2} - 1)$

Solution

Q.
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: [2026]

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

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

3 $2 - \sqrt{2}$

4 $\sqrt{2} - 1$

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Solution

Q. Let a1,a22,a322,…,a1029 be a G.P. of common ratio 12. If a1+a2+⋯+a10=62, then a1 is equal to: [2026]

1 2(2-2)

2 2(2-1)

3 2-2

4 2-1