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The value of the product $6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots$ up to
infinite terms is $\quad[2017-I I]$
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infinite terms is $\quad[2017-I I]$
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Verified Answer
The correct answer is:
6
$6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots .$ upto infinite terms.
$=6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots . \infty}=6$
Sum of $\infty$ terms of G.P. is $S_{\infty}=\frac{a}{1-r} .$ Here $a=\frac{1}{2}, r=\frac{1}{2}$
$\therefore 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=6^{\frac{1}{2}}=6^{1}=6$
Sum of $\infty$ terms ofG.P. is $S_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}} .$ Here $\mathrm{a}=\frac{1}{2}, \mathrm{r}=\frac{1}{2}$
$\therefore 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=6^{\frac{1}{2}}=6^{1}=6$
$=6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots . \infty}=6$
Sum of $\infty$ terms of G.P. is $S_{\infty}=\frac{a}{1-r} .$ Here $a=\frac{1}{2}, r=\frac{1}{2}$
$\therefore 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=6^{\frac{1}{2}}=6^{1}=6$
Sum of $\infty$ terms ofG.P. is $S_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}} .$ Here $\mathrm{a}=\frac{1}{2}, \mathrm{r}=\frac{1}{2}$
$\therefore 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=6^{\frac{1}{2}}=6^{1}=6$
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