There are many ways to solve this question. The basic idea is to assume that Charlotte makes a monthly payment of x and equate the amount she pays to the amount she owes, after adjusting both amounts with interest to be at the same point of time. In the following, we equate the amounts at the end of the term.
Initially, Charlotte owes $7680. She finishes her payments after a total of 6 + 36 = 42 months. Using a simple compounding formula, the amount she owes is worth P at the end of 42 months, where P is:
P = 7680 * (1 + .2045/12)^42 = 15616.67379
Now, the first installment she pays (at the end of six months) is paid 35 months in advance of the end, so it is worth x * (1 + .2375/12)^35 at the end of her loan period.
Similarly, the second installment is worth x * (1 + .2375/12)^34 at the end of the loan period.
Continuing, this way, the last installment is worth exactly x at the end of the loan period.
So, the total amount she paid equals:
x [(1 + .2375/12)^35 + (1 + .2375/12)^34 + ... + (1 + .2375/12)^0]
To calculate this, assume that 1+.2045/12 = a. Then the amount Charlotte pays is:
x (a^35 + a^34 + ... + a^0) = x (a^36 - 1)/(a - 1)
Clearly, this value must equal P, so we have:
x (a^36 - 1)/(a - 1) = P = 15616.67379
Substituting, a = 1 + .2045/12 and solving, we get
x = 317.82
