The 5th, 8th and 11th terms of a GP are a, b, c respectively

It is given in the question that $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of GP are $a, b$ and $c$ respectively.

Let us assume the GP is A, AR, AR $^{2}$, and AR3....

So, the $n^{\text {th }}$ term of this GP is $a_{n}=A R^{n-1}$

Now, $5^{\text {th }}$ term, $a_{5}=A R^{4}=a \rightarrow(1)$

$8^{\text {th }}$ term, $a_{8}=A R^{7}=b \rightarrow(2)$

$11^{\text {th }}$ term, $a_{11}=A R^{10}=c \rightarrow(3)$

Dividing equation (3) by (2) and (2) by (1),

$\frac{(3)}{(2)} \rightarrow \frac{\mathrm{AR}^{10}}{\mathrm{AR}^{7}}=\mathrm{R}^{3}=\frac{\mathrm{c}}{\mathrm{b}} \rightarrow(4)$

$\frac{(2)}{(1)} \rightarrow \frac{\mathrm{AR}^{7}}{\mathrm{AR}^{4}}=\mathrm{R}^{3}=\frac{\mathrm{b}}{\mathrm{a}} \rightarrow(5)$

So, both equation (4) and (5) gives the value of R3 . So we can equate them.

$\frac{c}{b}=\frac{b}{a}=R^{3}$

$\therefore b^{2}=a c$

Hence proved.