Find the geometric series whose

Solution:

The $n^{\text {th }}$ term of $a$ GP is $a_{n}=a r^{n-1}$

It's given in the question that $5^{\text {th }}$ term of the GP is 80 and $8^{\text {th }}$ term of GP is 640 .

So, $a_{5}=a r^{4}=80 \rightarrow(1)$

$a_{8}=a r^{7}=640 \rightarrow(2)$

$\frac{(2)}{(1)} \rightarrow \frac{\mathrm{ar}^{7}}{\mathrm{ar}^{4}}=\mathrm{r}^{3}=\frac{640}{80}=8$

Common ratio, r = 2,

$a r^{4}=80$

$16 a=80$

a = 5

The required GP is of the form $a, a r, a r^{2}, a r^{3}, a r^{4} \ldots .$

First term of GP, $a=5$

Second term of GP, ar $=5 \times 2=10$

Third term of GP, ar $^{2}=5 \times 2^{2}=20$

Fourth term of GP, $\operatorname{ar}^{3}=5 \times 2^{3}=40$

Fifth term of GP, $a r^{4}=5 \times 2^{4}=80$

And so on...

The required GP is $5,10,20,40,80 \ldots$