Q32 / 75 · JEE Main 2026 (04 April Shift 2) · Mathematics

Let $\alpha=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\ldots\infty$ and $\beta=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots\infty$. Then the value of $(0.2)^{\log_{\sqrt{5}}(\alpha)}+(0.04)^{\log_5(\beta)}$ is equal to:

  • A
    $4$
  • B
    $5$
  • Correct answer
    C
    $8$
  • D
    $25$

Solution

The sum of an infinite geometric progression is given by $S_{\infty} = \dfrac{a}{1-r}$.

For $\alpha$, $a = \dfrac{1}{4}$ and $r = \dfrac{1}{2}$:
$\alpha = \dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}} = \dfrac{1}{2}$

For $\beta$, $a = \dfrac{1}{3}$ and $r = \dfrac{1}{3}$:
$\beta = \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = \dfrac{1}{2}$

Now, evaluating the first term:
$(0.2)^{\log_{\sqrt{5}}(\alpha)} = \left(5^{-1}\right)^{\log_{5^{1/2}}(1/2)}$
Using the property $\log_{a^k}(x) = \dfrac{1}{k}\log_a(x)$:
$\log_{5^{1/2}}(1/2) = 2\log_5(1/2) = \log_5(1/4)$
Thus, $\left(5^{-1}\right)^{\log_5(1/4)} = 5^{-\log_5(1/4)} = 5^{\log_5(4)} = 4$

Evaluating the second term:
$(0.04)^{\log_5(\beta)} = \left(5^{-2}\right)^{\log_5(1/2)}$
$\left(5^{-2}\right)^{\log_5(1/2)} = 5^{-2\log_5(1/2)} = 5^{\log_5(4)} = 4$

Adding both terms:
$(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_5(\beta)} = 4 + 4 = 8$

Answer: $8$

Hence, the correct answer is option: 'C'.

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