The principal (P) is $6000 and the interest rate (r) is 0.02.

Now if the loan was for 1 month we would have

`L=P(1+r)` or `P=L/(1+r)`

If the loan was for 2 months we would have

`(1+r)((1+r)P-L)=L` or `(1+r)^2P-(1+r)L=L` which gives `P=L/(1+r)+L/(1+r)^2`

So if the loan is n months we would have

`P = sum_(j=1)^n L/(1+r)^j=L sum_(j=1)^n 1/(1+r)^n`

We can solve this by noting that

`(1+r)P=L sum_(j=0)^(n-1) 1/(1+r)^n`

So

`(1+r)P - P = rP = L(1-1/(1+r)^n)`

So `L=(rP)/(1-1/(1+r)^n)`

Substituting P = $6000, r = 0.02 and n = 24

###### `L = (0.02(6000))/(1-1/(1-0.02)^24)`

Evaluating we get $317.23 per month. Now this does seem high since 6000/24=250 but 2% per month is 48% for the entire loan, so this is actually a high interest rate made more attractive by stating it as a monthly interest rate.

So the final answer is $317.23 will pay off the loan in 24 months.