Determine the greatest value of $n$ for which $b > a$ Let $a_n$ and $b_n$ be two recursive sequences so that:
$a_{n+1} = a_{n} + 2000$ where $a_{1} = 500$
and
$b_{n+1} = \frac{b_{n}}{1,001}$ where $b_{1} = 50000$

Determine the greatest value of $n$ for which $b &gt; a$.

I have no idea of how to solve this. Perhaps the following is a start?

$a_{n+1} = a_{n} + 2000 \Rightarrow a_n = a_{n+1} - 2000$
$b_{n+1} = \frac{b_{n}}{1,001} \Rightarrow b_{n} = 1,001b_{n+1}$ 

And then you divide them and try to find a value for $n$ so that the fraction is greater than 1?
You should find a closed form for both sequences. I can tell you that
$$b_n = 5000 \cdot 1.001^{n-1} \\
a_n = 500 + 2000 \cdot (n-1) = 2000 n - 1500$$
Now you can solve $a_n = b_n$ for $n\in\mathbb R$ (an extension of the sequences). The resulting $n_0$ must then be rounded down for an answer to the question.
I believe you can simply work out a few cases by hand in a table for the answer.

Otherwise, find a simple formula for each term of the two sequences in this way:

Notice that $a_1 = 500, a_2 = 2000 + a_1, a_3 = 2000 + a_2 = 2 \times 2000 + a_1, \cdots$, which tells you that $a_n = 2000(n-1) + 500 = 2000n - 1500$.

Similarly, $b_n = \frac{b_1}{1001^{n-1}}$.

Now $$b_n &gt; a_n \\ \implies \frac{50000}{1001^{n-1}} &gt; 2000n - 1500 \\
\implies \frac{100}{1001^{n-1}} &gt; 5n - 3$$ and so on.