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 :

• A. $2 - \sqrt{2}$
• B. $2(2 - \sqrt{2})$
• C. $\sqrt{2} - 1$
• D. $2(\sqrt{2} - 1)$

Hint:

Recognize how the general term formula for a G.P. can relate two different sequences. Here, $a_n$ inherits the G.P. structure.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Let the terms of the G.P. be $b_n$. Then $b_n = \frac{a_n}{2^{n-1}}$.

Step 2: Detailed Explanation:
Given $b_n = b_1 \cdot r^{n-1}$ where $b_1 = a_1$ and $r = \frac{1}{\sqrt{2}}$.
\[ \frac{a_n}{2^{n-1}} = a_1 \left( \frac{1}{\sqrt{2}} \right)^{n-1} \implies a_n = a_1 \cdot 2^{n-1} \cdot \left( \frac{1}{\sqrt{2}} \right)^{n-1} \]
\[ a_n = a_1 \left( \frac{2}{\sqrt{2}} \right)^{n-1} = a_1 (\sqrt{2})^{n-1}. \]
Thus $a_n$ itself is a G.P. with first term $a_1$ and common ratio $R = \sqrt{2}$.
Sum of 10 terms:
\[ S_{10} = a_1 \frac{R^{10} - 1}{R - 1} = a_1 \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1} = a_1 \frac{32 - 1}{\sqrt{2} - 1} = a_1 \frac{31}{\sqrt{2} - 1}. \]
Given $S_{10} = 62$:
\[ a_1 \frac{31}{\sqrt{2} - 1} = 62 \implies \frac{a_1}{\sqrt{2} - 1} = 2 \implies a_1 = 2(\sqrt{2} - 1). \]

Step 3: Final Answer:
$a_1 = 2(\sqrt{2} - 1)$.