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


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

  • $\begingroup$ 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. $\endgroup$ – Tyler Mar 22 '15 at 19:57
  • $\begingroup$ I used the formula PV = A(i)/1-(1+(i)^-nt) as directed by my teacher, but it still isn't coming out correctly. $\endgroup$ – Hannah Mar 22 '15 at 20:02
  • $\begingroup$ 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. $\endgroup$ – Tyler Mar 22 '15 at 20:04
  • $\begingroup$ Thanks, you've been insanely helpful. $\endgroup$ – Hannah Mar 22 '15 at 20:05
  • $\begingroup$ 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) $\endgroup$ – Tyler Mar 22 '15 at 20:10

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).


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