Question

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 $b$, be equal to the $p$-th term of another GP, whose first term is $a$ and fifth term is $b$. Then $p$ is equal to

• A. 24
• B. 25
• C. 21
• D. 18

Answer 1

Answer: (C)

21

Answer 2

Define the first GP with first term $t_1 = a$ and common ratio $r_1$. Given $t_3 = b$, we have:

$t_3 = a \times r_1^2 = b \implies r_1 = \sqrt{\frac{b}{a}}.$

The 11th term $t_{11}$ of the first GP is:

$t_{11} = a \times r_1^{10} = a \times \left( \sqrt{\frac{b}{a}} \right)^{10} = \frac{b^5}{a^4}.$

Define the second GP with first term $T_1 = a$ and common ratio $r_2$. Given $T_5 = b$, we have:

$T_5 = a \times r_2^4 = b \implies r_2 = \left( \frac{b}{a} \right)^{\frac{1}{4}}.$

The p th term $T_p$ of the second GP is:

$T_p = a \times r_2^{p-1} = a \times \left( \frac{b}{a} \right)^{\frac{p-1}{4}}.$

Since $t_{11} = T_p$, we have:

$\frac{b^5}{a^4} = a \times \left( \frac{b}{a} \right)^{\frac{p-1}{4}}.$

Dividing both sides by $a$, we get:

$\frac{b^5}{a^5} = \left( \frac{b}{a} \right)^{\frac{p-1}{4}}.$

Equate the exponents:

$5 = \frac{p - 1}{4}.$

Solving for $p$:

$p - 1 = 20 \implies p = 21.$