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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Sequences and Series

Let a_ 1 , a_ 2 2 , a_ 3 2^ 2 , , a_ 10 2^ 9 be a G.P. of common...

Let a1,a22,a322,…,a1029a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \ldots, \frac{a_{10}}{2^{9}} be a G.P. of common ratio 12\frac{1}{\sqrt{2}}. If a1+a2+…+a10=62a_{1}+a_{2}+\ldots+a_{10}=62, then a1a_{1} is equal to :

Options

  1. A2−22-\sqrt{2}
  2. B2(2−2)2(2-\sqrt{2})
  3. C2−1\sqrt{2}-1
  4. D2(2−1)2(\sqrt{2}-1)

Correct answer

D. 2(2−1)2(\sqrt{2}-1)

Step-by-step solution

The G.P. is a1,a22,a322,…,a1029a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9} with common ratio 12\frac{1}{\sqrt{2}}.
So an2n−1=a1⋅(12)n−1\frac{a_n}{2^{n-1}} = a_1 \cdot \left(\frac{1}{\sqrt{2}}\right)^{n-1}, giving an=a1⋅(2)n−1a_n = a_1 \cdot (\sqrt{2})^{n-1}.
∑n=110an=a1⋅(2)10−12−1=a1⋅312−1=31a1(2+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: a1=22+1=2(2−1)a_1 = \frac{2}{\sqrt{2}+1} = 2(\sqrt{2}-1).
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