Sequences And Series Question 14
Question 14 - 2024 (30 Jan Shift 2)
Let $a$ and $b$ 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 $p$ is equal to
(1) 20
(2) 25
(3) 21
(4) 24
Show Answer
Answer (3)
Solution
$1^{\text {st }} GP \Rightarrow t _1=a, t _3=b=ar^{2} \Rightarrow r^{2}=\frac{b}{a}$
$t _{11}=ar^{10}=a\left(r^{2}\right)^{5}=a \cdot\left(\frac{b}{a}\right)^{5}$
$2^{\text {nd }}$ G. $P . \Rightarrow T _1=a, T _5=ar^{4}=b$
$\Rightarrow r^{4}=\left(\frac{b}{a}\right) \Rightarrow r=\left(\frac{b}{a}\right)^{1 / 4}$
$T _p=ar^{p-1}=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$
$t _{11}=T _p \Rightarrow a\left(\frac{b}{a}\right)^{5}=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$
$\Rightarrow 5=\frac{p-1}{4} \Rightarrow p=21$
