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JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Sequences and Series

Let

α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty

and

β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \infty

. 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✓ Correct
(D)25

Answer: (C)

Solution

The geometric series give

α = \frac{1/4}{1 - 1/2} = \frac{1}{2}

and

β = \frac{1/3}{1 - 1/3} = \frac{1}{2}

. For the first term,

0.2 = 5^{- 1}

and

lo g_{5} \frac{1}{2} = 2 lo g_{5} \frac{1}{2}

, so

(5^{- 1})^{2 l o g_{5} (1/2)} = (5^{l o g_{5} (1/2)})^{- 2} = (\frac{1}{2})^{- 2} = 4

. For the second,

0.04 = 5^{- 2}

, so

(5^{- 2})^{l o g_{5} (1/2)} = (\frac{1}{2})^{- 2} = 4

. The sum is

4 + 4 = 8

.

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option

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