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 \( p \):

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

Answer 2

To solve this problem, we need to analyze two geometric progressions (GPs) with given conditions, and find a relation between them. Let’s break down the steps:

1. Firstly, we define the first GP where the first term is given as \( a \) and the third term is \( b \). Let the common ratio of this GP be \( r_1 \). Thus, the terms of this GP can be expressed as: 
   \(T_1 = a\), \(T_3 = ar_1^2 = b\).
2. From \(T_3 = ar_1^2 = b\), we can solve for \( r_1 \): 
   \(r_1^2 = \frac{b}{a}\) hence, \(r_1 = \sqrt{\frac{b}{a}}\).
3. In the same GP, we are given that the 11th term is equal to the \( p \)-th term of another GP. The 11th term of the first GP is: 
   \(T_{11} = ar_1^{10} = a\left(\sqrt{\frac{b}{a}}\right)^{10} = a\left(\frac{b}{a}\right)^5 = \frac{b^5}{a^4}\).
4. Next, consider the second GP, whose first term is \( a \) and the fifth term is \( b \). Let the common ratio of this GP be \( r_2 \). The relevant terms are: 
   \(S_1 = a\), \(S_5 = ar_2^4 = b\).
5. Solving for \( r_2 \): 
   \(r_2^4 = \frac{b}{a}\) so, \(r_2 = \left(\frac{b}{a}\right)^{1/4}\).
6. The \( p \)-th term of the second GP should also be expressed as: 
   \(S_p = ar_2^{p-1} = ar_2^{p-1}\) To equate this to the 11th term of the first GP: 
   \(ar_2^{p-1} = \frac{b^5}{a^4}\).
7. Substitute \( r_2 = \left(\frac{b}{a}\right)^{1/4} \): 
   \(a\left(\frac{b}{a}\right)^{\frac{p-1}{4}} = \frac{b^5}{a^4}\).
8. Solve this equation: 
   \(\left(\frac{b}{a}\right)^{\frac{p-1}{4}} = \frac{b^5}{a^5}\).
9. Equating exponents: 
   \(\frac{p-1}{4} = 5\) which gives, \(p-1 = 20\), so, \(p = 21\).

Therefore, the value of \( p \) that satisfies both conditions is 21. Thus, the correct answer is 21.