Let a1, a22, a32^2,, a102^9 be a G.P. of common ratio 1 2. If a1 + a2 + + a10 = 62, then a1 is equal to:

Given: a1, a22, a32^2,, a102^9 is a G.P. with common ratio 1 2, and a1 + a2 + + a10 = 62. Find: a1. Let the terms of the G.P. be bn, where bn = an2^n-1 Since bn is a G.P. with first term b1 = a1 and common ratio 1 2, an2^n-1 = a1 ( 1 2)^n-1 Therefore, an = a1 2^n-1 ( 1 2)^n-1 an = a1 ( 2 2)^n-1 = a1( 2)^n-1 So an itself is a G.P. with first term a1 and common ratio 2. Now the sum of the first 10 terms is S10 = a1 ( 2)^10 - 1 2 - 1 Since ( 2)^10 = 32, S10 = a1 32 - 1 2 - 1 = a1 31 2 - 1 Given S10 = 62, a1 31 2 - 1 = 62 a1 2 - 1 = 2 a1 = 2( 2 - 1) Therefore, the correct option is D.

Let a1, a22, a32^2,, a102^9 be a G.P. of common ratio 1 2. If a1… — Solution | JEE Mathematics

MCQMediumJEE 2026Geometric Progression (GP)

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

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \dots, \frac{a_{10}}{2^9}$ is a G.P. with common ratio $\frac{1}{\sqrt{2}}$ , and $a_1 + a_2 + \dots + a_{10} = 62$ .

Find: $a_1$ .

Let the terms of the G.P. be $b_n$ , where

b_n = \frac{a_n}{2^{n-1}}

Since $b_n$ is a G.P. with first term $b_1 = a_1$ and common ratio $\frac{1}{\sqrt{2}}$ ,

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

Therefore,

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}

So $a_n$ itself is a G.P. with first term $a_1$ and common ratio $\sqrt{2}$ .

Now the sum of the first $10$ terms is

S_{10} = a_1 \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1}

Since $(\sqrt{2})^{10} = 32$ ,

S_{10} = 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

\frac{a_1}{\sqrt{2} - 1} = 2

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

Therefore, the correct option is D.

Common mistakes

Treating $a_n$ itself as the given G.P. from the start is incorrect because the G.P. is formed by $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \dots$ . First define $b_n = \frac{a_n}{2^{n-1}}$ , then relate $a_n$ to that sequence.
Using the common ratio as $\frac{1}{\sqrt{2}}$ for the sequence $a_n$ is wrong. After multiplying by $2^{n-1}$ , the ratio for $a_n$ becomes $\frac{2}{\sqrt{2}} = \sqrt{2}$ .
Applying the sum of G.P. formula with an incorrect power, such as using $R^9$ instead of $R^{10}$ for $10$ terms, gives the wrong result. For $10$ terms, use $S_{10} = a_1\frac{R^{10}-1}{R-1}$ .

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Answer

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