Question

The 4th and 7th terms of a G.P. are $\frac{1}{27}$ and $\frac{1}{792}$ respectively. Find the sum of n terms of the G.P.

Answer 1

$t_4=\frac{1}{27},t_7=\frac{1}{792},t_n=a{r}^{n-1}$

Where $t_n=n^{th}$ term, r = common difference, n = number of terms.

$t_4=a{r}^3=\frac{1}{27}$ ....(i)

$t_7=a{r}^6=\frac{1}{792}$ ....(ii)

Dividing (ii) by (i), we get

$\frac{t_7}{t_4}=\frac{a{r}^6}{a{r}^3}=r^3=\frac{27}{729}=\frac{1}{27}$,

$r=\frac{1}{3}$

Sum of n term $=S_n=\frac{a(1-r^n)}{1-r}$ .......(iii)

When, r = 3, $t_4=a{r}^3=\frac{1}{27}$

$a{\left(\frac{1}{3}\right)}^3=\frac{1}{27}$

$a=1$

Substituting $a=1,r=\frac{1}{3}in\left(iii\right)$

$S_n=\frac{1(1-{\left(\frac{1}{3}\right)}^n)}{1-\frac{1}{3}}$

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

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