Question

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

Answer 1

$4th$ term $=a{r}^3=\frac{1}{27}$

$7th$ term $=a{r}^6=\frac{1}{729}$

$\frac{{ar}^6}{{ar}^3}=\frac{1}{729}\times \frac{27}{1}\Rightarrow r^3=\frac{1}{27}\Rightarrow r=\frac{1}{3}{ar}^3=\frac{1}{27}\Rightarrow a\times \frac{1}{3^2}=\frac{1}{27}\Rightarrow \frac{a}{27}=\frac{1}{27}\Rightarrow a=1,r=\frac{1}{3}$

Sum of $n$ terms$=a\left(\frac{r^n-1}{r-1}\right)=a\left(\frac{1-r^n}{1-r}\right)=a\left(\frac{1-{\left(\frac{1}{r}\right)}^n}{1-\frac{1}{r}}\right)$

$=a\left[\frac{1-\frac{1}{3^n}}{1-\frac{1}{3}}\right]=1\times \left[\frac{3^n-1}{3^n}\right]\times \frac{1}{\frac{2}{3}}=\frac{3^n-1}{3^n}\times \frac{3}{2}=\frac{3(3^n-1)}{2\times 3^n}$