Sequences & SeriesmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Geometric Series with Log Exponents: Value = 8 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let α=14+18+116+⋯∞\alpha=\dfrac14+\dfrac18+\dfrac1{16}+\cdots\infty and β=13+19+127+⋯∞\beta=\dfrac13+\dfrac19+\dfrac1{27}+\cdots\infty. Then the value of (0.2)log⁡5(α)+(0.04)log⁡5(β)(0.2)^{\log_{\sqrt5}(\alpha)}+(0.04)^{\log_5(\beta)} is
A44
B55
C88correct
D2525
Solution
Step 1: Sum the geometric series:
α=1/41−12=12,β=1/31−13=12.\alpha=\frac{1/4}{1-\frac12}=\frac12,\qquad \beta=\frac{1/3}{1-\frac13}=\frac12.
 Step 2: First term. Since 0.2=5−1=(5)−20.2=5^{-1}=(\sqrt5)^{-2}, use alog⁡bc=clog⁡baa^{\log_b c}=c^{\log_b a}:
(0.2)log⁡5α=αlog⁡50.2=α−2=(12)−2=4.(0.2)^{\log_{\sqrt5}\alpha}=\alpha^{\log_{\sqrt5}0.2}=\alpha^{-2}=\left(\frac12\right)^{-2}=4.
 Step 3: Second term. Since 0.04=5−20.04=5^{-2}:
(0.04)log⁡5β=5−2log⁡5β=β−2=(12)−2=4.(0.04)^{\log_5\beta}=5^{-2\log_5\beta}=\beta^{-2}=\left(\frac12\right)^{-2}=4.
 Step 4: Sum:
4+4=8.4+4=8.
 Correct answer: (3)
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