Thread: Monthly Payments + Interest Problem

Hello!

I have a problem that I'm trying to figure out how to do the steps of. I have the solution, but I can't figure out how to get to it. The problem is as follows:

"Gina wants to establish a college fund for her newborn daughter that will have accumulated $120,000 at the end of 18 yr. If she can count on an interest rate of 6%, compounded monthly, how much should she deposit each month to accomplish this?"

The answer is $309.79, but I don't see how. Can someone please explain the steps to me? Thanks in advance.

There's a formula for that sort of thing, but feel free to build it from basic principles.

n = 18 years = 18*12 months = 216 months

i = 0.06

i^{(12)} = 0.06/12 = 0.005

Monthly Accumulation Factor: a = 1+i^{(12)} = 1.005

Monthly Deposit: P <== This is what we seek.

Final Accumulation: A = $120,000

If you had to do it in one payment, what would it look like?

P*a = 120000 and P = 120000/1.005 = 119402.99

If you had to do it in two payments, what would it look like?

= P(1.010025 + 1.005) = P(2.015025) and P = 120000/2.015025 = 59.552.62

It is hoped that you can see into the past and answer this last question, what would it look like if you have 216 months?



Your task is to add up all that stuff in the parentheses. It is a finite geometric series. Can you finish?

I get $308.26, so perhaps we have not defined the problem correctly.

The formula for these types of problems is as follows

For the balance at the time of the last deposit:

Future Value = (C/r)[((1+r)^n) - 1]

Where C is your deposit, n is the number of deposits and r is the relevant interest rate.

C = ?
r = 0.06/12 = 0.005
n = 12 x 18 = 216
FV = 120000

120000 = (C/0.005)[(1.005^216) - 1]

Solve for C and you get $309.79.

As for not using the formula, you can think of the 120000 as the future value of all the deposits. I.e., the first deposit will have a value of C(1.005)^216, 2nd will have a value of C(1.005)^215, etc... Add them all up and it will total 120000.

It can be expressed as a sum of a geometric series. Use the formula

S = (a(r^n - 1))/(r - 1)

with S = 120000
r = 1.005
n = 216

Solve for A and you get 309.79

Thanks so much, Tiger_shark! I spent all afternoon trying to figure that out. The book I have didn't list anything remotely close to that formula, so it was killing me to figure out what it was. Thanks once again!