Question

The $5^{th},8^{th}$ and ${11}^{th}$ terms of a G.P. are $p,q$ and $s$ respectively. Show that $q^2=ps$.

Answer 1

Let $a$ be the first term and $r$ be the common ratio of G.P.
According to the given condition,
$a_5=a{r}^{5-1}=a{r}^4=p...\left(1\right)a_8=a{r}^{8-1}=a{r}^7=q...\left(2\right)a_{11}=a{r}^{11-1}=a{r}^{10}=s...\left(3\right)$
Dividing equation (2) by (1), we obtain
$\frac{{ar}^7}{{ar}^4}=\frac{q}{p}\Rightarrow r^3=\frac{q}{p}....\left(4\right)$
Dividing equation (3) by (2), we obtain
$\frac{{ar}^{10}}{{ar}^7}=\frac{s}{q}\Rightarrow r^3=\frac{s}{q}....\left(5\right)$
From (4) and (5) we obtain
$\frac{q}{p}=\frac{s}{q}\Rightarrow q^2=ps$