Question

The sixth and tenth terms of a Geometric Progression are $63$ and $5103$ respectively. Find the first term and common ratio.

Answer 1

Given : $a_6=63$ and $a_{10}=5103$

$n^{th}$ term of a Geometric Progression is given by $a_n=a{r}^{n-1}$, where $a$ is the first term and $r$ is the common ratio.

$\therefore$ $a_6=a{r}^5=63$ ..... $\left(1\right)$ and $a_{10}=a{r}^9=5103$ ....... $\left(2\right)$

Dividing equation $\left(2\right)$ by $\left(1\right)$ we get,

$\frac{a{r}^9}{a{r}^5}=\frac{5103}{63}$

$⟹$ $r^4=81$

$\therefore$ $r=3$ ............... taking fourth root on both sides

Substitute the value of r in equation $\left(1\right)$ we get,

$a(3{)}^5=63$

$⟹$ $a\times 243=63$

$⟹$ $a=\frac{7}{27}$

$\therefore$ first term is $a=\frac{7}{27}$ and common ratio $r=3$