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

To solve this problem, we need to identify the first term \(a_1\) of the given geometric progression (G.P.) and verify it using the provided condition that the sum of a series of terms equals 62.

The given series is: 

\(a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}\)

This can be rewritten as:

\(a_1, \frac{a_2}{2}, \frac{a_3}{4}, \ldots, \frac{a_{10}}{512}\)

We know this is a G.P. with a common ratio of \(\frac{1}{\sqrt{2}}\). Let's determine the actual terms:

• \(a_2 = a_1 \cdot \frac{1}{\sqrt{2}}\), then \(\frac{a_2}{2} = \frac{a_1}{2 \sqrt{2}}\)
• \(a_3 = a_2 \cdot \frac{1}{\sqrt{2}} = a_1 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{a_1}{2}\), then \(\frac{a_3}{4} = \frac{a_1}{8}\)

Continuing this pattern, it becomes more apparent that:

\(a_n = a_1 \left(\frac{1}{\sqrt{2}}\right)^{n-1}\)

The sequence can therefore be structured as:

\(a_1, \frac{a_1}{2 \sqrt{2}}, \frac{a_1}{8}, \ldots\)

Summing it up, we have:

\(a_1 \left( 1 + \frac{1}{2\sqrt{2}} + \frac{1}{8} + \cdots \right) = 62\)

It's evident that the series is geometric when redesigned as \(a_1 \text{G.P.}\), giving the formula:

\(S = \frac{a_1 (1 - r^n)}{1 - r}\)

Given:

\(r = \frac{1}{\sqrt{2}} \left(\frac{1}{2}\right) = \frac{1}{\sqrt{2^3}} = \frac{1}{2\sqrt{2}}\)

The sum of the G.P. is provided to be 62:

\(\frac{a_1 (1 - \left(\frac{1}{2\sqrt{2}}\right)^{10})}{1 - \frac{1}{2\sqrt{2}}} = 62\)

This results in solving for:

\(a_1 = 2(\sqrt{2} - 1)\)

Therefore, the correct answer is:

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