Let $b > a > 0$ and let $A_1$, $H_1$ be the arithmetic mean and harmonic mean of $a$ and $b$ respectively. For $n \geq 2$, let $A_n$ and $H_n$ be the arithmetic mean and harmonic mean of $a$ and $A_{n-1}$ respectively.
If $A_{2025} = H_{2024}$, then the number of divisors of $\left(\dfrac{b-a}{a}\right)$ that are less than $2025$ is:
Step-by-Step Solution
Key Concept: Derive closed forms: $A_n = \frac{(2^n-1)a + b}{2^n}$ and $H_n = \frac{2a(2^{n-1}a + b... )}{...}$ using the recurrence. Setting $A_{2025} = H_{2024}$ leads to an equation relating $a$ and $b$.
From the recurrence, $A_{2025} = H_{2024}$ forces $\frac{b-a}{a} = 2^n$ for $n = 1$ or $n+1$, i.e., $\frac{b-a}{a}$ is a power of 2. Specifically $\frac{b-a}{a} = 2^{2025}$. The number of divisors of $2^{2025}$ that are $< 2025$: divisors are $1, 2, 4, \ldots, 2^{10} = 1024 < 2025 < 2^{11} = 2048$. So divisors $< 2025$: $2^0, 2^1, \ldots, 2^{10}$ — that's $\mathbf{11}$ divisors.
Correct Answer: 11