Q41 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \ldots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_{1}+a_{2}+\ldots+a_{10}=62$, then $a_{1}$ is equal to :
- A$2-\sqrt{2}$
- B$2(2-\sqrt{2})$
- C$\sqrt{2}-1$
- Correct answerD$2(\sqrt{2}-1)$
Solution
The G.P. is $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ with common ratio $\frac{1}{\sqrt{2}}$.
So $\frac{a_n}{2^{n-1}} = a_1 \cdot \left(\frac{1}{\sqrt{2}}\right)^{n-1}$, giving $a_n = a_1 \cdot (\sqrt{2})^{n-1}$.
$\sum_{n=1}^{10} a_n = a_1 \cdot \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1} = a_1 \cdot \frac{31}{\sqrt{2} - 1} = 31a_1(\sqrt{2} + 1)$.
Setting this equal to 62: $a_1 = \frac{2}{\sqrt{2}+1} = 2(\sqrt{2}-1)$.
So $\frac{a_n}{2^{n-1}} = a_1 \cdot \left(\frac{1}{\sqrt{2}}\right)^{n-1}$, giving $a_n = a_1 \cdot (\sqrt{2})^{n-1}$.
$\sum_{n=1}^{10} a_n = a_1 \cdot \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1} = a_1 \cdot \frac{31}{\sqrt{2} - 1} = 31a_1(\sqrt{2} + 1)$.
Setting this equal to 62: $a_1 = \frac{2}{\sqrt{2}+1} = 2(\sqrt{2}-1)$.
Hence, the correct answer is option: 'D'.