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The sum of the series formed by the sequence $3, \sqrt{3}, 1 \ldots \ldots$ upto infinity is :
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Verified Answer
The correct answer is:
$\frac{3 \sqrt{3}(\sqrt{3}+1)}{2}$
$3, \sqrt{3}, 1, \frac{1}{\sqrt{3}}, \ldots \ldots, \infty$
This is a Geometric Progression with $\mathrm{a}=3, \mathrm{r}=\frac{1}{\sqrt{3}}$
$\begin{aligned} \mathrm{S}_{\infty} &=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{3}{1-\frac{1}{\sqrt{3}}} \\ &=\frac{3 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3 \sqrt{3}(\sqrt{3}+1)}{2} \end{aligned}$
This is a Geometric Progression with $\mathrm{a}=3, \mathrm{r}=\frac{1}{\sqrt{3}}$
$\begin{aligned} \mathrm{S}_{\infty} &=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{3}{1-\frac{1}{\sqrt{3}}} \\ &=\frac{3 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3 \sqrt{3}(\sqrt{3}+1)}{2} \end{aligned}$
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