Question

If the 4th and 9th terms of a GP are 54 and 13122 respectively, find the GP. Also, find its general term.

Answer 1

Let a be the first term and r be the common ratio of the given GP.

Then, $T_4=54\Rightarrow a{r}^{(4-1)}=54\Rightarrow a{r}^3=54$ . . .. (i)

And, $T_9=13122\Rightarrow a{r}^{(9-1)}=13122\Rightarrow a{r}^8=13122$ . . .(ii)

On dividing (ii) by (i), we get

$\frac{a{r}^8}{a{r}^3}=\frac{13122}{54}\Rightarrow r^5=243=3^5\Rightarrow r=3$

Putting r = 3 in (i), we get

$a\times 3^3=54\Rightarrow 27a=54\Rightarrow a=2$

Thus, a = 2 and r = 3.

So, the required GP is 2, 6, 18, 54, ... .

The general term of the GP is given by

$T_n=a{r}^{(n-1)}=\{2\times 3^{(n-1)}\}$