Let a and b be two distinct positive real numbers. Let the 11th term of a GP, whose first term is a and third term is b , is equal to p t h term of another GP, whose first term is a and fifth term is b . Then p is equal to [2024]

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Solution

Q.
Let $a$ and $b$ be two distinct positive real numbers. Let the 11th term of a GP, whose first term is $a$ and third term is $b,$ is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b .$ Then $p$ is equal to [2024]

1 21

2 24

3 20

4 25

Ans.
(1)

Given, $T_{1} = a, T_{3} = a r_{1}^{2} = b$

$\Rightarrow r_{1} = {(\frac{b}{a})}^{1 / 2}$ ...(i)

Also, $T_{1} = a, T_{5} = b$

$\Rightarrow a r_{2}^{4} = b \Rightarrow r_{2} = {(\frac{b}{a})}^{1 / 4}$ ...(ii)

And $11^{t h}$ term of first GP = $p^{t h}$ term of second GP

Now, $a r_{1}^{10} = a r_{2}^{p - 1}$

$\Rightarrow a {(\frac{b}{a})}^{5} =$ $a$ ${{(\frac{b}{a})}^{1 / 4}}^{p - 1}$ (Using (i) and (ii))

$\Rightarrow {(\frac{b}{a})}^{5} = {(\frac{b}{a})}^{\frac{p - 1}{4}} \Rightarrow 5 = \frac{p - 1}{4} \Rightarrow p - 1 = 20 \Rightarrow p = 21$

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