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

To solve the problem, we need to understand that both \( A \) and \( B \) are geometric series. We can write the given series in standard geometric form:
\[ A = 1 + r^a + r^{2a} + r^{3a} + \dots \]This is an infinite geometric series with the first term \( T_1 = 1 \) and a common ratio \( r^a \). The sum of an infinite geometric series is given by the formula:
\[ S = \frac{T_1}{1-r} \]Thus, for series \( A \), we have:
\[ A = \frac{1}{1-r^a} \](assuming \( |r^a| < 1 \))
Similarly, convert series \( B = 1 + r^b + r^{2b} + r^{3b} + \dots \) into the standard form:
\[ B = \frac{1}{1-r^b} \](assuming \( |r^b| < 1 \))
Now, we equate \( \frac{a}{b} \) to an expression involving \( A \) and \( B \). Using the relationship¹:
\(\frac{1-r^a}{1-r^b} = \frac{B(A-1)}{A(B-1)}\)
Taking the logarithm on both sides simplifies to:
\(\frac{a}{b} = \log_\frac{b-1}{b} \left(\frac{A-1}{A}\right)\)
Therefore, the correct answer is \( \log_\frac{b-1}{b} \left(\frac{A-1}{A}\right) \). This involves expressing the ratio \( \frac{a}{b} \) in terms of the series sums \( A \) and \( B \) by appropriately choosing the logarithmic form that ties these expressions together.

Answer 2

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).