Question

The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

Answer 1

General term $=a{r}^{n-1}$

where, n=nth term

r=common ratio

a=first term

Now, $a{r}^3=\mathrm{10..............}\left(1\right).a{r}^6=\mathrm{80............}\left(2\right)a{r}^{n-1}=\mathrm{2560.................}\left(3\right)$

Dividing (2) by (1)

$r^3=8\Rightarrow r=2$

in (1)

$a\times 8=10\Rightarrow a=\frac{10}{8}$

in (3)

$\frac{10}{8}\times 2^{n-1}=2560\Rightarrow 2^{n-1}=256\Rightarrow 2^{n-1}=2^8\Rightarrow n-1=8\Rightarrow n=9$