Let a and b be two distinct positive real numbers. Let 11 th term of a GP, whose first term is a and third…

Given,

$1^{st} GP\Rightarrow \text{First term}=a, a_3=b$

$\Rightarrow a{r}^2=b$

$\Rightarrow r^2=\frac{b}{a}$

$\Rightarrow a_{10}=a{r}^{10}=a{\left(\frac{b}{a}\right)}^5 ...\left(i\right)$

$2^{nd} GP\Rightarrow \text{First term}=a, a_5=b$

$\Rightarrow a{R}^4=b ...\left(ii\right)$

$\Rightarrow a_p=a{R}^{p-1}$

$\Rightarrow a_p=a{\left[{\left(\frac{b}{a}\right)}^{\frac{1}{4}}\right]}^{p-1} ...\left(ii\right)$

Now, using the given condition of $p^{th}$ term we get,

$\Rightarrow a{\left(\frac{b}{a}\right)}^5=a{\left(\frac{b}{a}\right)}^{\frac{p-1}{4}}$

$\Rightarrow {\left(\frac{b}{a}\right)}^5={\left(\frac{b}{a}\right)}^{\frac{p-1}{4}}$

$\Rightarrow 5=\frac{p-1}{4}$

$\Rightarrow p=21$