Question

Show that the progressions are GP. Also, find the common ratio.

$a,\frac{3a^2}{4},\frac{9a^3}{16},.....$

Answer 1

Examine the progressions are GP and find the common ratio:

The sequence is $a,\frac{3a^2}{4},\frac{9a^3}{16},.....$

Here, $a_1=a,a_2=\frac{3a^2}{4},a_3=\frac{9a^3}{16}$

$\Rightarrow \frac{a_2}{a_1}=\frac{3a^2}{4\left(a\right)}=\frac{3a}{4}$

$\Rightarrow \frac{a_3}{a_2}=\frac{9a^3}{16}\times \frac{4}{3a^2}=\frac{3a}{4}$ and so on.

Since the ratio of any term by its preceding term is constant.

Hence, the given sequence is GP and the common ratio $r=\frac{3a}{4}$.