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 $\mathrm{p}$ is equal to
(1) 20
(2) 25
(3) 21
(4) 24
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Answer (3)
Solution
$1^{\text {st }} \mathrm{GP} \Rightarrow \mathrm{t}{1}=\mathrm{a}, \mathrm{t}{3}=\mathrm{b}=\mathrm{ar}^{2} \Rightarrow \mathrm{r}^{2}=\frac{\mathrm{b}}{\mathrm{a}}$
$\mathrm{t}_{11}=\mathrm{ar}^{10}=\mathrm{a}\left(\mathrm{r}^{2}\right)^{5}=\mathrm{a} \cdot\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{5}$
$2^{\text {nd }}$ G. $P . \Rightarrow \mathrm{T}{1}=\mathrm{a}, \mathrm{T}{5}=\mathrm{ar}^{4}=\mathrm{b}$
$\Rightarrow \mathrm{r}^{4}=\left(\frac{\mathrm{b}}{\mathrm{a}}\right) \Rightarrow \mathrm{r}=\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{1 / 4}$
$\mathrm{T}_{\mathrm{p}}=\mathrm{ar}^{\mathrm{p}-1}=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{\mathrm{p}-1}{4}}$
$\mathrm{t}{11}=\mathrm{T}{\mathrm{p}} \Rightarrow \mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{5}=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{\mathrm{p}-1}{4}}$
$\Rightarrow 5=\frac{\mathrm{p}-1}{4} \Rightarrow \mathrm{p}=21$
