The value of y -0-36A- 0-9- 0-8C- 0-6D- 0-25

The correct option is C 0.6Given: y = (0.36)^log_0.25( 1 3 +13^2+13^3+…∞) Now, we know that, 1 3 +13^2+13^3+…∞ nbsp;= 1 3 1 13 = 12 = 0.5 Now, the given ...

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Standard XII

Mathematics

Sum of n Terms

The value of ...

Question

The value of $y = (0.36)^{{log}_{0.25} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots \infty)}$ is

0.9

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0.8

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0.6

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0.25

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Solution

The correct option is C $0.6$
Given:
$y = (0.36)^{{log}_{0.25} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots \infty)}$
Now, we know that,
$\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots \infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2} = 0.5$
Now, the given expression can be rewritten as,
$y = (0.36)^{{log}_{0.25} 0.5} \Rightarrow y = (0.6)^{2 {log}_{0.25} 0.5} \Rightarrow y = (0.6)^{{log}_{0.25} 0.25} ∴ y = 0.6$

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Sum of n Terms

Standard XII Mathematics

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The value of y=(0.36)log0.25(13+132+133+…∞) is

The correct option is C 0.6Given: y=(0.36)log0.25(13+132+133+…∞) Now, we know that, 13+132+133+…∞=131−13=12=0.5 Now, the given expression can be rewritten as, y=(0.36)log0.250.5⇒y=(0.6)2log0.250.5⇒y=(0.6)log0.250.25∴y=0.6

The value of $y = (0.36)^{{log}_{0.25} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots \infty)}$ is