Question

How many terms of the sequence $\sqrt{3},3,3\sqrt{3},$ ... must be taken to make the sum $39+13\sqrt{3}$?

Answer 1

Here, a = $\sqrt{3}$
Common ratio, r = $\sqrt{3}$
Sum of n terms, S_n = $39+3\sqrt{3}$

$$\begin{gathered} S_n=\sqrt{3}\left(\frac{{\left(\sqrt{3}\right)}^n-1}{\sqrt{3}-1}\right) \\ \Rightarrow 39+13\sqrt{3}=\frac{\sqrt{3}}{\left(\sqrt{3}-1\right)}\left\{{\left(\sqrt{3}\right)}^n-1\right\} \\ \Rightarrow {\left(\sqrt{3}\right)}^n-1=\frac{\left(39+13\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}} \\ \Rightarrow {\left(\sqrt{3}\right)}^n=1+26 \\ \Rightarrow {\left(\sqrt{3}\right)}^n=27 \\ \Rightarrow {\left(\sqrt{3}\right)}^n={\left(\sqrt{3}\right)}^6 \\ \therefore \mathit{}n=6 \end{gathered}$$