Question

If \( A = 1 + r^a + r^{2a} + r^{3a} + \dots \infty \) and \( B = 1 + r^b + r^{2b} + r^{3b} + \dots \infty \), then \( \frac{a}{b} \) is equal.

• A. \( \log_{b}(A) \)
• B. \( \log_{1-b}(1 - A) \)
• C. \( \log_\frac{b-1}{b} \left(\frac{A-1}{A}\right) \)
• D. None of these

Hint:

For problems involving geometric series and logs, transform the series into their sum formula first, then apply logarithmic identities.

Answer 1

The Correct Option is C

We are given the following two infinite geometric series: 1. For \( A \): \[ A = \frac{1}{1 - r^a} \] This implies that: \[ 1 - r^a = \frac{1}{A} \] Rearranging this: \[ r^a = 1 - \frac{1}{A} \] Thus, we can write: \[ r^a = \frac{A - 1}{A} \] 2. For \( B \): \[ B = \frac{1}{1 - r^b} \] This implies that: \[ 1 - r^b = \frac{1}{B} \] Rearranging this: \[ r^b = 1 - \frac{1}{B} \] Thus, we can write: \[ r^b = \frac{B - 1}{B} \] Now, to solve for \( \frac{a}{b} \), we take the logarithm of both sides of each equation. \[ a \log r = \log \left( \frac{A - 1}{A} \right) \] \[ b \log r = \log \left( \frac{B - 1}{B} \right) \] Next, we divide the two equations to find \( \frac{a}{b} \): \[ \frac{a}{b} = \frac{\log \left( \frac{A - 1}{A} \right)}{\log \left( \frac{B - 1}{B} \right)} \] Thus, we have: \[ \frac{a}{b} = \log_r \left( \frac{A - 1}{A} \right) = \log_r \left( \frac{B - 1}{B} \right) \] Thus, the answer matches option (C).