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Sequences and Series question

2020 · 3 Sep · Shift 1 · Q40
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Sequences and Series question

2020 · 3 Sep · Shift 1 · Q40

JEE MainMathematicsSequences and SeriesNumerical+4 / −1
The value of (0.16)log⁡2.5(13+132+....to ∞){\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}(0.16)log2.5​(31​+321​+....to∞) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4

  1. Evaluate the infinite series

The expression inside the logarithm is

13+132+133+⋯\frac13 + \frac1{3^2} + \frac1{3^3} + \cdots31​+321​+331​+⋯

This is an infinite geometric series with:

a=13,r=13a = \frac13, \quad r = \frac13a=31​,r=31​

Since ∣r∣<1|r|<1∣r∣<1, its sum is

S=a1−r=131−13=1323=12S = \frac{a}{1-r} = \frac{\frac13}{1-\frac13} = \frac{\frac13}{\frac23} = \frac12S=1−ra​=1−31​31​​=32​31​​=21​

So the given expression becomes

(0.16)log⁡2.5(1/2)(0.16)^{\log_{2.5}(1/2)}(0.16)log2.5​(1/2)
  1. Rewrite the numbers as fractions
0.16=16100=4250.16 = \frac{16}{100} = \frac{4}{25}0.16=10016​=254​

Also,

2.5=522.5 = \frac522.5=25​

Hence,

(425)log⁡5/2(1/2)\left(\frac{4}{25}\right)^{\log_{5/2}(1/2)}(254​)log5/2​(1/2)

Now observe that

425=(25)2=(15/2)2=(25)2\frac{4}{25} = \left(\frac25\right)^2 = \left(\frac{1}{5/2}\right)^2 = \left(\frac{2}{5}\right)^2254​=(52​)2=(5/21​)2=(52​)2

So the expression is

((25)2)log⁡5/2(1/2)=(25)2log⁡5/2(1/2)\left(\left(\frac25\right)^2\right)^{\log_{5/2}(1/2)} = \left(\frac25\right)^{2\log_{5/2}(1/2)}((52​)2)log5/2​(1/2)=(52​)2log5/2​(1/2)

But

25=(52)−1\frac25 = \left(\frac52\right)^{-1}52​=(25​)−1

Therefore,

((52)−1)2log⁡5/2(1/2)=(52)−2log⁡5/2(1/2)\left(\left(\frac52\right)^{-1}\right)^{2\log_{5/2}(1/2)} = \left(\frac52\right)^{-2\log_{5/2}(1/2)}((25​)−1)2log5/2​(1/2)=(25​)−2log5/2​(1/2)

Using the identity alog⁡ax=xa^{\log_a x}=xaloga​x=x,

(52)log⁡5/2(1/2)=12\left(\frac52\right)^{\log_{5/2}(1/2)} = \frac12(25​)log5/2​(1/2)=21​

Hence,

(52)−2log⁡5/2(1/2)=(12)−2=4\left(\frac52\right)^{-2\log_{5/2}(1/2)} = \left(\frac12\right)^{-2} = 4(25​)−2log5/2​(1/2)=(21​)−2=4
  1. Final answer
4\boxed{4}4​
  1. Comparison with stored answer

Stored correct answer = 444.

This matches the derived answer.

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