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Let $\mathrm{a}$ and $\mathrm{b}$ be be two distinct positive real numbers. Let $11^{\text {th }}$ term of a $GP$, whose first term is $a$ and third term is $b$, is equal to $p^{\text {th }}$ term of another $GP$, whose first term is $a$ and fifth term is $b$. Then $\mathrm{p}$ is equal to
$20$
$25$
$21$
$24$
Solution
$ 1^{\text {st }} G P \Rightarrow t_1=a, t_3=b=a r^2 \Rightarrow r^2=\frac{b}{a} $
$ 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 =a 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$
