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 p^th term of another GP, whose first term is a and fifth term is b. Then p is equal to

1^ st GP => t_1= a, t_3= b= ar^2 => r^2=frac b a t_11=a r^10=a(r^2)^5=a * ((b / a))^5 2^ nd G.P. => T_1=a, T_5=a r^4=b => r^4=((b / a)) => r=((b / a))^1 / 4 T_p=r^p-1=a((b / a))^(p-1 / 4) t_11=T_p => a((b / a))^5=a((b / a))^(p-1 / 4) => 5=(p-1 / 4) => p=21 \

Let a and b be two distinct positive real numbers. Let ... | Sequence & Series (Mathematics)

$\begin{aligned} & 1^{\text {st }} \mathrm{GP} \Rightarrow \mathrm{t}_1=\mathrm{a}, \mathrm{t}_3=\mathrm{b}=\mathrm{ar}^2 \Rightarrow \mathrm{r}^2=\frac{\mathrm{b}}{\mathrm{a}} \\& \mathrm{t}_{11}=a r^{10}=a\left(r^2\right)^5=a \cdot\left(\frac{b}{a}\right)^5 \\& 2^{\text {nd }} \text { G.P. } \Rightarrow T_1=a, T_5=a r^4=b \\& \Rightarrow r^4=\left(\frac{b}{a}\right) \Rightarrow r=\left(\frac{b}{a}\right)^{1 / 4} \\& T_p=r^{p-1}=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}} \\& t_{11}=T_p \Rightarrow a\left(\frac{b}{a}\right)^5=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}} \\& \Rightarrow 5=\frac{p-1}{4} \Rightarrow p=21 \end{aligned}$

Explanation

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