Question

The 4th and 7th terms of a G.P. are $\frac{1}{27}\mathrm{and}\frac{1}{729}$ respectively. Find the sum of n terms of the G.P.

Answer 1

Let a be the first term and r be the common ratio of the G.P.

$$\begin{gathered} \therefore a_4=\frac{1}{27} \\ \Rightarrow a{r}^{4-1}=\frac{1}{27} \\ \Rightarrow a{r}^3=\frac{1}{27} \\ \Rightarrow {\left(a{r}^3\right)}^2=\frac{1}{{27}^2} \\ \Rightarrow a^2{r}^6=\frac{1}{729} \\ \Rightarrow a{r}^6=\frac{1}{729a}...\left(\mathrm{i}\right) \\ \mathrm{Similarly},{\mathrm{a}}_7=\frac{1}{729} \\ \Rightarrow a{r}^{7-1}=\frac{1}{729} \\ \Rightarrow a{r}^6=\frac{1}{729} \\ \Rightarrow a{r}^6=\frac{1}{729\mathrm{a}}\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \therefore a=1 \\ \mathrm{Putting}\mathrm{this}\mathrm{in}{a}_4=\frac{1}{27} \\ \Rightarrow a{r}^3=\frac{1}{3^3} \\ \Rightarrow r^3=\frac{1}{3^3} \\ \therefore r=\frac{1}{3} \\ \mathrm{Now},\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{G}.\mathrm{P}.,S_n=a\left(\frac{r^n-1}{r-1}\right) \\ \Rightarrow S_n=1\left(\frac{1-{\left({\displaystyle \frac{1}{3}}\right)}^n}{1-{\displaystyle \frac{1}{3}}}\right) \\ \Rightarrow S_n=\frac{3}{2}\left(1-\frac{1}{3^n}\right) \end{gathered}$$