1. ## Find Total Payment

A loan of 2400 dollars is to be paid in the following manner; 300 dollars will be paid at the end of each year plus 10% interest on the unpaid balance at the beginning of the year. Find the total of all the payments.

a. 5160
b. 4256
c. 5120
d. 6267

My Work:

The loan is for $2400. Let's see: 2400/300 = 8. It will take 8 years to make the full payment plus 10 percent on the unpaid balance. 300 x 0.10 = 30. Then 300 + 30 = 330. The total of all payments should be 330 + 2400 or$2730.

My answer is not close to the choices given. Help....

2. ## Re: Find Total Payment

I can't be sure, but I think you may have misunderstood the problem. Please give the problem statement exactly and completely

3. ## Re: Find Total Payment

Basic arithmetic sequence and series.
Originally Posted by nycmath
It will take 8 years to make the full payment plus 10 percent on the unpaid balance.
300 x 0.10 = 30.
The unpaid balance at the beginning of the 1st year is 2400.
Thus, total payment at the end of the 1st year is 300 + 2400(.10)
The unpaid balance at the beginning of the 2nd year is 2400-300 and so on and so forth.
By this reckoning, none of the choices given are correct.
But my present state of stupor (alcohol and pain meds) and debilitating stiff neck could be making me see things that are not really there.

4. ## Re: Find Total Payment

2400 * 10% = 240 + 300 = 540
2100 * 10% = 210 + 300 = 510
1800 * 10% = 180 + 300 = 480
1500 * 10% = 150 + 300 = 450
1200 * 10% = 120 + 300 = 420
900 * 10% = 90 + 300 = 390
600 * 10% = 60 + 300 = 360
300 * 10% = 30 + 300 = 330
total payment = 3480

5. ## Re: Find Total Payment

Originally Posted by hbahrami
2400 * 10% = 240 + 300 = 540
This should be 2400 * 10% + 300 = 240 + 300 = 540
The same goes for the rest.
You should have let the OP do the all the work. Before you know it, you'll get sucked into the often thankless business of saving the world with all this math knight errantry.
A compact expression might be (you'll have to check, my beer goggles are a bit foggy):
$\sum\limits_{t = 1}^{8}$300+[2400-300(t-1)](.10)