Answer:
B. [tex]4\sqrt{3}-6[/tex]
Step-by-step explanation:
We have,
The first term of the series, [tex]a=\sqrt{3}[/tex].
The common difference is given by, [tex]r=\frac{\frac{-3}{2}}{\sqrt{3}}[/tex] i.e. [tex]r=\frac{-\sqrt{3}}{2}[/tex].
Since, the given series is an infinite series, then,
Sum of an infinite series = [tex]\frac{a}{1-r}[/tex]
i.e. Sum the series = [tex]\frac{\sqrt{3}}{1+\frac{\sqrt{3}}{2}}[/tex]
i.e. Sum the series = [tex]\frac{2\sqrt{3}}{2+\sqrt{3}}[/tex]
i.e. Sum the series = [tex]\frac{2\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}[/tex]
i.e. Sum the series = [tex]\frac{2\sqrt{3}\times (2-\sqrt{3})}{4-3}[/tex]
i.e. Sum the series = [tex]4\sqrt{3}-6[/tex]
Thus, the sum of the series is [tex]4\sqrt{3}-6[/tex].
