Let α = 1/4 +1/8 +1/16 +ldots ∞ and β = 1/3 +1/9 +1/27 +ldots ∞ . Then (0.2) log 5 (α) + (0.04) log 5 (β) is equal to :

Mathematics · Sequence and Series

JEE Main 2026 — 4 April, Evening Shift — Question 31

Let $\alpha = \frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\ldots \infty$ and $\beta = \frac{1}{3} +\frac{1}{9} +\frac{1}{27} +\ldots \infty$ . Then $(0.2)^{\log_{5}(\alpha)} + (0.04)^{\log_{5}(\beta)}$ is equal to :

Option A:
4
Option B:
5
Option C:
8
Correct
Option D:
25

Answer: C

Step-by-step solution

$\alpha=\frac{1 / 4}{1-\frac{1}{2}}=\frac{1}{2}$ $\beta=\frac{1 / 3}{1-\frac{1}{3}}=\frac{1}{2}$ $(0.2)^{\log _{\sqrt{5}} \alpha}=(\alpha)^{\log _{\sqrt{5}} \frac{1}{5}}=\left(\frac{1}{2}\right)^{-2}=4$ $(0.04)^{\log _{5} \beta}=5^{-2 \log _{5} \beta}=\left(\frac{1}{2}\right)^{-2}=4$ $\Rightarrow 4+4=8$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Geometric Progression