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JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Sequences and Series

The value of

(0.16)^{l o g_{2.5} (\frac{1}{3} + \frac{1}{3 ^{2}} + \frac{1}{3 ^{3}} + \dots to \infty)}

is equal to ______.

Answer: 4

Solution

The geometric series sums to

\frac{1/3}{1 - 1/3} = \frac{1}{2}

, so the expression is

(0.16)^{l o g_{2.5} (1/2)}

. Writing

0.16 = \frac{4}{25} = (\frac{5}{2})^{- 2}

, it becomes

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

.

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Integer answer

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