# Monthly Payments + Interest Problem

• September 12th 2007, 06:45 PM
audax
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.
• September 13th 2007, 05:33 AM
TKHunny
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?

$P*a^{2} + P*a = 120000 = P(a^{2} + a)$ = 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?

$P(a^{216} + a^{215} + a^{214} + ... + a^{2} + a) = 120000$

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.
• September 13th 2007, 01:46 PM
Tiger_shark
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
• September 13th 2007, 04:37 PM
audax
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!