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The least value of n in order that the sum of first n terms of an infinite series \( \Large 1 + \frac{3}{4} + \left(\frac{3}{4}\right)^{2} + \left(\frac{3}{4}\right)^{3}+ .... \) should differ from the sum of the series by less then \( \Large 10^{-6} \) is
[\( \Large\ Given\ log_{10}2=0.30103, log_{10}3=0.47712 \)]
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A) 14 |
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B) 27 |
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C) 53 |
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D) 57 |
Correct Answer:
| C) 53 |
Description for Correct answer:
Series \( \Large 1+\frac{3}{4}+ \left(\frac{3}{4}\right)^{2}+...+nterms=\frac{1- \left(\frac{3}{4}\right)^{n} }{1-\frac{3}{4}} \)
and \( \Large 1+\frac{3}{4}+ \left(\frac{3}{4}\right)^{2}+...=\frac{1}{1-\frac{3}{4}}=4 \)
Now, \( \Large \frac{1- \left(\frac{3}{4}\right)^{n} }{\frac{1}{4}}=4-10^{-6} \)
=> \( \Large 1- \left(\frac{3}{4}\right)^{n}=1-\frac{1}{4} \left(10^{-6}\right) \)
=> \( \Large \left(\frac{3}{4}\right)^{n}=\frac{10^{-6}}{4} \)
=> \( \Large nlog_{10}\frac{3}{4}=log_{10}10^{-6}-log_{10}4 \)
=> \( \Large n \left(0.47712-2 \times 0.30103\right)=-6-2 \times \left(0.30103\right) \)
=> \( \Large n=\frac{6.6026}{0.12494}=53 \)
Part of solved Logarithms questions and answers : >> Elementary Mathematics >> Logarithms
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