Let a and b be two distinct positive real num | vedclass

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(C) For the $1^{	ext{st}}$ $GP$: Let the common ratio be $r_1$. Given $t_1 = a$ and $t_3 = b = a r_1^2$,so $r_1^2 = \frac{b}{a}$.The $11^{	ext{th}}$

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) For the $1^{	ext{st}}$ $GP$: Let the common ratio be $r_1$. Given $t_1 = a$ and $t_3 = b = a r_1^2$,so $r_1^2 = \frac{b}{a}$.The $11^{	ext{th}}$ term is $t_{11} = a r_1^{10} = a (r_1^2)^5 = a \left(\frac{b}{a}ight)^5$.For the $2^{	ext{nd}}$ $GP$: Let the common ratio be $r_2$. Given $T_1 = a$ and $T_5 = b = a r_2^4$,so $r_2^4 = \frac{b}{a}$,which implies $r_2 = \left(\frac{b}{a}ight)^{1/4}$.The $p^{	ext{th}}$ term is $T_p = a r_2^{p-1} = a \left(\frac{b}{a}ight)^{\frac{p-1}{4}}$.Given $t_{11} = T_p$,we have $a \left(\frac{b}{a}ight)^5 = a \left(\frac{b}{a}ight)^{\frac{p-1}{4}}$.Equating the exponents,$5 = \frac{p-1}{4}$,which gives $p-1 = 20$,so $p = 21$.

Let $a$ and $b$ be two distinct positive real numbers. Let the $11^{\text{th}}$ term of a $GP$,whose first term is $a$ and third term is $b$,be equal to the $p^{\text{th}}$ term of another $GP$,whose first term is $a$ and fifth term is $b$. Then $p$ is equal to:

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