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UPSC NDA Math Model Paper 3

Question : 30 of 120

Marks: +1, -0

6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \dots

up to infinite terms is

Solution:

Given that:

6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \dots

upto infinite terms

\Rightarrow 6\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots\right)

\dots (\because a^{m} \times a^{n}=a^{m+n})

The series:

\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots

is in form of infinite geometric series.

Comparing it with standard infinite G.P series, we get a = 1/2 and r = 1/2

Sum of infinite terms of a G.P,

s_{n}=\frac{a}{1-r}

\Rightarrow s_{n}=\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)

\Rightarrow s_{n}=\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)=1

\Rightarrow 6\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots\right)=6^{1}=6

Hence, the value of product of

6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \dots

upto infinite terms = 6

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