Question

Find the sum of the geometric series. a b c d

Answer 1

Answer:

B. $4\sqrt{3}-6$

Step-by-step explanation:

We have,

The first term of the series, $a=\sqrt{3}$.

The common difference is given by, $r=\frac{\frac{-3}{2}}{\sqrt{3}}$ i.e. $r=\frac{-\sqrt{3}}{2}$.

Since, the given series is an infinite series, then,

Sum of an infinite series = $\frac{a}{1-r}$

i.e. Sum the series = $\frac{\sqrt{3}}{1+\frac{\sqrt{3}}{2}}$

i.e. Sum the series = $\frac{2\sqrt{3}}{2+\sqrt{3}}$

i.e. Sum the series = $\frac{2\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

i.e. Sum the series = $\frac{2\sqrt{3}\times (2-\sqrt{3})}{4-3}$

i.e. Sum the series = $4\sqrt{3}-6$

Thus, the sum of the series is $4\sqrt{3}-6$.