Let α = 1/4 + 1/8 + 1/16 + ... ∞ and β = 1/3 + 1/9 + 1/27 + ... ∞. Then the value of (0.2)^log_sqrt(5)(α) + (0.04)^log_4(β) is equal to:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q7 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #394

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)^{\log_{\sqrt{5}} (α)} + (0.04)^{\log_{4} (β)}

is equal to:

Clue 1 of 3: Identify the Series

The sums for $α$ and $β$ represent infinite geometric progressions.

🥷

Ninja Strategy•Symmetry and Typo Detection

Recognize that the first term is 4. The second term's structure is similar, and assuming a typo from $\log_{4}$ to $\log_{5}$ for symmetry makes the second term also 4, leading to the answer 8.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate the values of α and β✦ Active

Both $α$ and $β$ are sums of infinite geometric progressions (GP). We use the formula for the sum of an infinite GP, $S_{\infty} = \frac{a}{1 - r}$ .

For $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ , the first term is $a = \frac{1}{4}$ and the common ratio is $r = \frac{1}{2}$ .

α = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}

For $β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$ , the first term is $a = \frac{1}{3}$ and the common ratio is $r = \frac{1}{3}$ .

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

Step 2: Evaluate the first term of the expression○ Expand

Substitute $α = 1 / 2$ into the first term, $(0.2)^{\log_{\sqrt{5}} (α)}$ , and simplify using logarithm properties. Note that $0.2 = 1 / 5 = 5^{- 1}$ and $\sqrt{5} = 5^{1 / 2}$ .

(0.2)^{\log_{\sqrt{5}} (1 / 2)} = (5^{- 1})^{\log_{5^{1 / 2}} (2^{- 1})} = (5^{- 1})^{\frac{- 1}{1 / 2} \log_{5} (2)} = (5^{- 1})^{- 2 \log_{5} (2)} = 5^{2 \log_{5} (2)} = 5^{\log_{5} (2^{2})} = 4

Alternatively, using the property $a^{\log_{b} c} = c^{\log_{b} a}$ :

(0.2)^{\log_{\sqrt{5}} (1 / 2)} = (1 / 2)^{\log_{\sqrt{5}} (0.2)} = (1 / 2)^{\log_{5^{1 / 2}} (1 / 5)} = (1 / 2)^{\log_{5^{1 / 2}} (5^{- 1})} = (1 / 2)^{\frac{- 1}{1 / 2} \log_{5} (5)} = (1 / 2)^{- 2} = 4

Step 3: Evaluate the second term and find the total value○ Expand

Substitute $β = 1 / 2$ into the second term, $(0.04)^{\log_{4} (β)}$ . A direct calculation gives $(0.04)^{\log_{4} (1 / 2)} = (1 / 25)^{\log_{2^{2}} (2^{- 1})} = (1 / 25)^{- 1 / 2} = 25^{1 / 2} = 5$ . This leads to a total sum of $4 + 5 = 9$ , which is not among the options. This suggests a likely typo in the question.

Assuming the base of the logarithm in the second term was intended to be 5 instead of 4, to create symmetry with the first term (since $0.04 = 1 / 25 = 5^{- 2}$ ), we evaluate $(0.04)^{\log_{5} (β)}$ .

(0.04)^{\log_{5} (1 / 2)} = (5^{- 2})^{\log_{5} (2^{- 1})} = (5^{- 2})^{- \log_{5} (2)} = 5^{2 \log_{5} (2)} = 5^{\log_{5} (2^{2})} = 4

With this correction, the total value is the sum of the two terms:

Total value = 4 + 4 = 8

💡 Teacher's Secret Hint

The question as written yields an answer of 9. The provided solution assumes a common type of exam question error where the base of the logarithm in the second term should have been 5 instead of 4. This correction leads to the answer 8, which is one of the options.

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Maths Question 7 – JEE-MAIN 2026

Clue 1 of 3: Identify the Series

🧠 Full Solution Path