Question

Let \( a \) and \( b \) be two distinct positive real numbers. Let the 11^th 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

The Correct Option is C

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 pth 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 \( p \):

\[ p - 1 = 20 \implies p = 21. \]