Mathematics · Sequence and Series

JEE Main 2026 — 24 January, Morning Shift — Question 4

Let $729,81,9,1, \ldots$ be a sequence and $\mathrm{P}_{\mathrm{n}}$ denote the product of the first $n$ terms of this sequence If $2 \sum_{n=1}^{40}\left(P_{n}\right)^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}}$ and $\operatorname{gcd}(\alpha, \beta)=1$ , then $\alpha+\beta$ is equal to

Option A:
$73$
Correct
Option B:
$74$
Option C:
$75$
Option D:
$76$

Answer: A

Step-by-step solution

$\mathrm{P}_{\mathrm{n}}=729.81 .9 \ldots \ldots$ . (n terms)

$=3^{6} \cdot 3^{4} \cdot 3^{2} \ldots \ldots \cdot 3^{-2 n+8}$

$P_{n}=3^{6+4+2+\ldots+(-2 n+8)}=3^{n(7-n)}$

$P_{n}^{1 / n}=3^{7-n}$

$\Rightarrow \sum_{\mathrm{n}=1}^{40}\left(\mathrm{P}_{\mathrm{n}}\right)^{\frac{1}{\mathrm{n}}}=3^{6}+3^{5}+\ldots . .+(40$ terms $)$

$=3^{6}\left[\frac{1-\left(\frac{1}{3}\right)^{40}}{1-\frac{1}{3}}\right]$

$=\frac{3^{6}\left[3^{40}-1\right] \times 3^{1}}{3^{40} \times 2}$

$\sum\left(\mathrm{P}_{\mathrm{n}}\right)^{\frac{1}{\mathrm{n}}}=\frac{\left(3^{40}-1\right)}{2 \times 3^{33}}, \quad \alpha=40$ $\beta=33$ $\alpha+\beta=73$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Telescopic Summation

Consider an A.P.: a 1, a 2, ldots, a n ; a 1 0 . If a 2- a 1 =-3/4, a n=1/4 a 1 , and sum i=1^ n a i=525/2 , then sum i=1^17 a i is equal…

Answer: A

Step-by-step solution

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