# The 4th and 7th terms of a GP are

Question:

The $4^{\text {th }}$ and $7^{\text {th }}$ terms of a GP are $\frac{1}{27}$ and $\frac{1}{729}$ respectively. Find the sum of $n$ terms of the GP.

Solution:

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

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

Dividing the $7^{\text {th }}$ term by the $4^{\text {th }}$ term,

$\frac{\mathrm{ar}^{6}}{\mathrm{ar}^{3}}=\frac{\frac{1}{729}}{\frac{1}{27}}$

$\Rightarrow r^{3}=\frac{1}{27}$ ……(i)

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

$a r^{3}=\frac{1}{27}$ [putting from eqn (i) ]

$a^{\frac{1}{27}}=\frac{1}{27}$

$\therefore a=1$

Sum of a G.P. series is represented by the formula, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}$

when |r|<1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 1

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

n terms

$\mathrm{S}_{\mathrm{n}}=1 \times \frac{1-\frac{1}{3}^{\mathrm{n}}}{1-\frac{1}{3}}$

⇒  $\mathrm{S}_{\mathrm{n}}=\frac{1-\frac{1}{3}^{\mathrm{n}}}{\frac{2}{3}}$

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

$\therefore \mathrm{S}_{\mathrm{n}}=\frac{3-\frac{1}{2^{\mathrm{n}-1}}}{2}$