MCQ
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
  • A
    $20$
  • B
    $25$
  • $21$
  • D
    $24$

Answer

Correct option: C.
$21$
c
$ 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$

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