JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

Let $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ and \beta = \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \ldots\. Then the value of $(0.2)^{l o g_{5} (α)} + (0.04)^{l o g_{5} (β)}$ is equal to:

A
4
B
5
C
8
D
25

Solution

Both series are geometric. Sum $α = \frac{1/4}{1 - 1/2} = \frac{1}{2}$ . Sum $β = \frac{1/3}{1 - 1/3} = \frac{1}{2}$ .

First term: $(0.2)^{l o g_{5} (1/2)}$ . Note $0.2 = 1/5 = 5^{- 1}$ , and $lo g_{5} (1/2) = 2 lo g_{5} (1/2) = - 2 lo g_{5} 2$ . So $(0.2)^{- 2 l o g_{5} 2} = 5^{2 l o g_{5} 2} = 2^{2} = 4$ .

Second term: $(0.04)^{l o g_{5} (1/2)} = (5^{- 2})^{- l o g_{5} 2} = 5^{2 l o g_{5} 2} = 4$ .

Sum $= 4 + 4 = 8$ .

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