# Can't Get Present Value Answer?

I've done this problem at least 20 times a number of different ways, but I can't seem to get the correct answer. Please show all work and describe the EXACT formula you used:

Find the present value of the ordinary annuity: Payments of $78 are made quarterly for 10 years at 8% compounded quarterly. Answer:$2133.73

Thanks!

I used the formula PV = A(i)/1-(1+(i)^-nt) as directed by my teacher, but it still isn't coming out correctly.

• Can you tell us what you did 1 or 2 of those 20 times? If you understand how an annuity is calculated, the solution is immediate, this isn't a tricky problem. – Tyler Mar 22 '15 at 19:57
• I used the formula PV = A(i)/1-(1+(i)^-nt) as directed by my teacher, but it still isn't coming out correctly. – Hannah Mar 22 '15 at 20:02
• Well then, it certainly seems as if the issue is that you are using the formula as directed instead of understanding how and why such a formula exists. I suggest writing down a cash flow diagram, the associated present value of the payments, and realizing you're summing a geometric series. – Tyler Mar 22 '15 at 20:04
• Thanks, you've been insanely helpful. – Hannah Mar 22 '15 at 20:05
• My only other pieces of advice are to make sure you are using the correct interest rate, and double checking your formula. 8% compounded quarterly is 2% effective quarterly, and I've never seen present value described as you have laid it out (see e.g. wikipedia) – Tyler Mar 22 '15 at 20:10

## 1 Answer

Let's go through the basics. You are receiving an annuity of $\$78$each quarter for$40$quarters. The interest rate is$8\%$compounded quarterly, which is$2\%$effective quarterly. Let$P_i$denote the present value of the$i^{th}$payment.$P_1 = P(1+i)^{-1} = 78(1.02)^{-1}P_2 = P(1+i)^{-2} = 78(1.02)^{-2}$...$P_{40} = P(1+i)^{-40} = 78(1.02)^{-40}$The present value of the annuity is the sum of the above payments.$\sum_{i = 1}^{40} P_i = 78\sum_{i = 1}^{40} (\frac{1}{1.02})^i = \frac{78}{1.02}\sum_{i = 0}^{39} (\frac{1}{1.02})^i = \frac{78}{1.02}\frac{1 - \frac{1}{(1.02)^{40}}}{1-\frac{1}{1.02}}$where the$2^{\mathrm{nd}}\$ equality is just to rewrite the series as the familiar geometric series and the final equality is applying the closed form solution for a geometric series solution.

Again, I'd recommend trying to understand the steps of the solution so you can understand why there is a formula for such equations and what it is, otherwise you can be led astray by wrong formulas (or even misusing correct formulas).