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

Given,

$${1}^{st}GP\Rightarrow \text{First term}=a,{a}_{3}=b$$1stGP⇒First term=a,a3​=b

$$\Rightarrow a{r}^{2}=b$$⇒ar2=b

$$\Rightarrow {r}^{2}=\frac{b}{a}$$⇒r2=ab​

$$\Rightarrow {a}_{10}=a{r}^{10}=a{(\frac{b}{a})}^{5}...(i)$$⇒a10​=ar10=a(ab​)5...(i)

$${2}^{nd}GP\Rightarrow \text{First term}=a,{a}_{5}=b$$2ndGP⇒First term=a,a5​=b

$$\Rightarrow a{R}^{4}=b...(ii)$$⇒aR4=b...(ii)

$$\Rightarrow {a}_{p}=a{R}^{p-1}$$⇒ap​=aRp−1

$$\Rightarrow {a}_{p}=a{[{(\frac{b}{a})}^{\frac{1}{4}}]}^{p-1}...(ii)$$⇒ap​=a[(ab​)41​]p−1...(ii)

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

$$\Rightarrow a{(\frac{b}{a})}^{5}=a{(\frac{b}{a})}^{\frac{p-1}{4}}$$⇒a(ab​)5=a(ab​)4p−1​

$$\Rightarrow {(\frac{b}{a})}^{5}={(\frac{b}{a})}^{\frac{p-1}{4}}$$⇒(ab​)5=(ab​)4p−1​

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

$$\Rightarrow p=21$$⇒p=21