Let a and b be two distinct positive real numbers. Let 11 ^ th term...

Given,

${1}^{st}GP ⇒ \text{First term} = a, {a}_{3} = b$

$⇒ a{r}^{2} = b$

$⇒ {r}^{2} = \dfrac{b}{a}$

$⇒ {a}_{10} = a{r}^{10} = a{\left(\dfrac{b}{a}\right)}^{5} . . . \left(i\right)$

${2}^{nd}GP ⇒ \text{First term} = a, {a}_{5} = b$

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

$⇒ {a}_{p} = a{R}^{p - 1}$

$⇒ {a}_{p} = a{\left[\left(\dfrac{b}{a}\right)\dfrac{1}{4}\right]}^{p - 1} . . . \left(ii\right)$

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

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

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

$⇒ 5 = \dfrac{p - 1}{4}$

$⇒ p = 21$