JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

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 $\frac{1}{2}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to :

A
$2 (2 - 1)$
B
$2 - 2$
C
$2 - 1$
D
$2 (2 - 2)$

Solution

The ratio of consecutive terms of the given G.P. is $\frac{a _{k + 1} / 2 ^{k}}{a _{k} / 2 ^{k - 1}} = \frac{a _{k + 1}}{2 a _{k}} = \frac{1}{2}$ , so $\frac{a _{k + 1}}{a _{k}} = 2$ .

So $a_{1}, a_{2}, \dots, a_{10}$ is itself a G.P. with first term $a_{1}$ and common ratio $2$ .

Sum: $a_{1} \cdot \frac{( 2 ) ^{10} - 1}{2 - 1} = 62$ , i.e. $a_{1} \cdot \frac{32 - 1}{2 - 1} = 62$ .

$a_{1} = \frac{62 ( 2 - 1 )}{31} = 2 (2 - 1)$ .

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