JEE Main 2026 — Mathematics Algebra

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

For $\alpha$α, $a = \dfrac{1}{4}$a=41​ and $r = \dfrac{1}{2}$r=21​:

$$\alpha = \dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}} = \dfrac{1}{2}$$α=1−21​41​​=21​

For $\beta$β, $a = \dfrac{1}{3}$a=31​ and $r = \dfrac{1}{3}$r=31​:

$$\beta = \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = \dfrac{1}{2}$$β=1−31​31​​=21​

Now, evaluating the first term:

$$(0.2)^{\log_{\sqrt{5}}(\alpha)} = \left(5^{-1}\right)^{\log_{5^{1/2}}(1/2)}$$(0.2)log5​​(α)=(5−1)log51/2​(1/2)

Using the property $\log_{a^k}(x) = \dfrac{1}{k}\log_a(x)$logak​(x)=k1​loga​(x):

$$\log_{5^{1/2}}(1/2) = 2\log_5(1/2) = \log_5(1/4)$$log51/2​(1/2)=2log5​(1/2)=log5​(1/4)

Thus, $\left(5^{-1}\right)^{\log_5(1/4)} = 5^{-\log_5(1/4)} = 5^{\log_5(4)} = 4$(5−1)log5​(1/4)=5−log5​(1/4)=5log5​(4)=4

Evaluating the second term:

$$(0.04)^{\log_5(\beta)} = \left(5^{-2}\right)^{\log_5(1/2)}$$(0.04)log5​(β)=(5−2)log5​(1/2)

$$\left(5^{-2}\right)^{\log_5(1/2)} = 5^{-2\log_5(1/2)} = 5^{\log_5(4)} = 4$$(5−2)log5​(1/2)=5−2log5​(1/2)=5log5​(4)=4

Adding both terms:

$$(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_5(\beta)} = 4 + 4 = 8$$(0.2)log5​​(α)+(0.04)log5​(β)=4+4=8

Answer: $8$8