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

Consider a geometric progression (GP) with the first term \( t_1 = a \) and common ratio \( r_1 \). Given the third term \( t_3 = b \), the common ratio is determined as \( r_1 = \sqrt{\frac{b}{a}} \). The 11th term of this GP is \( t_{11} = a \times r_1^{10} = \frac{b^5}{a^4} \).

Now, consider a second GP with the first term \( T_1 = a \) and common ratio \( r_2 \). Given the fifth term \( T_5 = b \), the common ratio is \( r_2 = \left( \frac{b}{a} \right)^{\frac{1}{4}} \). The \( p \)-th term of this second GP is \( T_p = a \times r_2^{p-1} = a \times \left( \frac{b}{a} \right)^{\frac{p-1}{4}} \).

We are given that \( t_{11} = T_p \). Equating the expressions for these terms:

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

Simplifying the equation by dividing both sides by \( a \):

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

Equating the exponents of \( \frac{b}{a} \) on both sides:

\[5 = \frac{p - 1}{4}.\]

Solving for \( p \):

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