Question

In a $GP,t_{p+q}=a$ and $t_{p-q}=b$. Prove that $t_p=\sqrt{ab}$

Answer 1

Let the first term of the G.P. is $s$ and $r$ be the common ratio.

Also given,

$t_{p+q}=a$ and $t_{p-q}=b$

or, $s{r}^{p+q-1}=a$......(1) and $s{r}^{p-q-1}=b$......(2).

Now multiplying (1) and (2) we get,

$s^2{r}^{2p-2}=ab$

or, $s{r}^{p-1}=\sqrt{ab}$

or, $t_p=\sqrt{ab}$.