Question

Given an A.P. and a G.P. with positive terms, with the first and second terms of the progressions being equal. If \(a_n\) and \(b_n\) be the \(n\)-th term of A.P. and G.P. respectively, then:

• A. \(a_n>b_n\) for all \(n>2\)
• B. \(a_n<b_n\) for all \(n>2\)
• C. \(a_n = b_n\) for some \(n>2\)
• D. \(a_n = b_n\) for some odd \(n\)

Hint:

In problems involving A.P. and G.P., remember that the terms in an A.P. increase linearly, whereas the terms in a G.P. increase exponentially (when r > 1).

Answer 1

The Correct Option is B

Step 1: The general term of an A.P. is given by:

\[ a_n = a_1 + (n - 1)d \]

where \( a_1 \) is the first term and \( d \) is the common difference.

Step 2: The general term of a G.P. is given by:

\[ b_n = b_1 r^{n-1} \]

where \( b_1 \) is the first term and \( r \) is the common ratio.

Step 3: We are given that the first and second terms of both progressions are equal, i.e.,

\[ a_1 = b_1 \quad \text{and} \quad a_2 = b_2. \]

- For A.P., \( a_2 = a_1 + d \). For G.P., \( b_2 = b_1 r \).

Equating these:

\[ a_1 + d = b_1 r \]

Since \( a_1 = b_1 \), we get:

\[ d = b_1 (r - 1) \]

Step 4: Now, for \( n > 2 \), the general terms \( a_n \) and \( b_n \) are:

\[ a_n = a_1 + (n - 1)d = b_1 + (n - 1)b_1(r - 1) = b_1 [1 + (n - 1)(r - 1)] \] \[ b_n = b_1 r^{n-1} \]

Step 5: To compare \( a_n \) and \( b_n \), let’s examine their behavior for large \( n \):

• As \( n \) increases, the term \( (n - 1)(r - 1) \) will make \( a_n \) grow linearly, whereas \( b_n \) will grow exponentially (since \( r > 1 \) in a G.P.).
• Therefore, for \( n > 2 \), \( a_n \) will always be less than \( b_n \) for the G.P., hence \( a_n < b_n \).