Question

In a geometric sequence, the first term is $\frac{1}{3}$ and the sixth term is $\frac{1}{729}$, find the G.P.

Answer 1

It is given that the $1^{st}$ term of an G.P is $a=\frac{1}{3}$ and the $6^{th}$ term is $T_6=\frac{1}{729}$

We know that the general term of an geometric progression with first term $a$ and common ratio $r$ is $T_n=a{r}^{n-1}$, therefore,

$T_6=a{r}^{6-1}\Rightarrow \frac{1}{729}=a{r}^5.......\left(1\right)$

Substitute the value of $a$ in equation 1 as follows:

$\frac{1}{729}=a{r}^5\Rightarrow \frac{1}{729}=\frac{1}{3}{r}^5\Rightarrow r^5=\frac{3}{729}\Rightarrow r^5=\frac{1}{243}$

$\Rightarrow r^5=\frac{1^5}{3^5}\Rightarrow r^5={\left(\frac{1}{3}\right)}^5\Rightarrow r=\frac{1}{3}$

We also know the general term of G.P is $a,ar,a{r}^2,.....$, Therefore, the terms of G.P are as follows:

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

$ar=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

$a{r}^2=\frac{1}{3}\times {\left(\frac{1}{3}\right)}^2=\frac{1}{3}\times \frac{1}{9}=\frac{1}{27}$ and so on.

Hence, the G.P is $\frac{1}{3},\frac{1}{9},\frac{1}{27},.........$