JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ . Then $a_{3} + a_{5} + a_{7}$ is equal to :

A
3256
B
3252
C
3244
D
3248

Solution

Let first term $a$ and common ratio $r$ . Then $a_{2} a_{3} a_{4} = (a r) (a r^{2}) (a r^{3}) = a^{3} r^{6} = 64$ , so $a r^{2} = 4$ .

$a_{1} + a_{3} + a_{5} = a (1 + r^{2} + r^{4}) = \frac{813}{7}$ . Using $a = 4/ r^{2}$ , solving gives $r^{2} = 28$ .

$a_{3} + a_{5} + a_{7} = a r^{2} (1 + r^{2} + r^{4}) = 4 (1 + 28 + 784) = 4 \cdot 813 = 3252$ .

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