# Finance Test: How to solve for interest rate? (Annuities)

Hello thanks for helping me!

Here's the problem:

You are told that if you invest \$11,000 per year for 23 years (all payments made at the end of each year), you will have accumulated \$366,000 at the end of the period. What annual rate of return is the investment offering?

This is how far I've come with the problem:

`366000/11000 = ((1+r)^23 - 1)/r`

I don't know what to do next.

Thank you!

• This question cannot be solved by hand. You need a financial calculator. Commented Dec 12, 2021 at 11:04
• You can try from 0.1, 0.2, 0.3 etc then binary search. Commented Dec 12, 2021 at 11:29
• @JoãoDuarte If I asked you to calculate the \$√2\$ by hand, would you say "there's got to be a way"? There's lots of mathematical problems which are hard to solve.
– Tim
Commented Dec 12, 2021 at 21:23
• @Tim Manual square root is easy... not so much geometric series of course :)
– Joe
Commented Dec 13, 2021 at 5:05
• Commented Dec 13, 2021 at 5:55

This is actually a geometric series and r can not be solved for manually (easily).

As one of the formulas on the Wikipedia page I provided a link to just above implies, the sum of the terms in a geometric series of the form below can be calculated by

``````t^(N-1) + t^(N-2) + ... + t^0 = (1 - t^N)/(1-t).
``````

First, let's first generalize the problem you have shared slightly to make use of the formula above. Suppose one invests P amount of money at the end of every year for N years and every year's investment earns an interest rate of r annually. Then, the total Q to be accumulated at the end of the year N would be

``````Q   = P(1+r)^(N-1) + P(1+r)^(N-2) + ... + P(1+r)^(0)

= P[(1+r)^(N-1) + (1+r)^(N-2) + ... + 1]

Q/P = [(1+r)^(N-1) + (1+r)^(N-2) + ... + 1]

= [1 - (1+r)^N]/(1 - (1+r)) (using the geometric series summation formula above assuming t = 1+r)

= [(1+r)^N - 1]/r.          (after rearranging the terms and a little simplification)
``````

We can easily adapt this formula to the problem you have posted by assuming Q = \$366,000, P = \$11,000, N = 23 and r, of course, is the annual rate of return. Then

``````366,000 / 11,000 = [(1+r)^23 - 1]/r
``````

which is the same as the equation you came up with.

This equation can not be simplified meaningfully any further in order to find r directly. It is a root-finding problem for a polynomial of the 23rd degree. You need to use a numerical method such as the Newton-Rhapson method to solve it (which is how it is usually done) or you can go by trial-end-error manually in a systematic fashion. Another tool that can be used is the Solver add-in in Excel.

Of course, another way to go about finding r is to use a financial calculator which would have a built-in function for such problems and all you would need to do is to provide the Q, P, and N to the calculator. Yet another way is to use the RATE() function in Excel in a similar fashion to the financial calculator. Note that both the financial calculators and the RATE() function in Excel actually define the problem as a polynomial root-finding problem and use a numerical method such as the Newton-Rhapson behind-the-scenes.

Having said all that, the answer to your question is `r = 3.21%`.

Editing notice

Formulas and equations are revised to account for the fact that money is invested at the end of every year. Thx for the warning @base64.

• Nice explanation. The 1st payment is made at the end of the 1st year, so the return of the 1st payment is P(1+r)^(N-1). And the last payment has no return. His formula is correct. The answer is r = 0.0320855 calculator.net/… Commented Dec 12, 2021 at 14:26
• @Alper I believe you have a typo. You wrote: and r can not be solved for manually (easily). I think you meant: to write now instead of the word not.
– Bob
Commented Apr 24 at 2:14
• @Bob Thx but there is no typo. Commented Apr 24 at 9:55

You can try guesses for `r` until you get close enough to zero, i.e.

``````(11000 ((1 + r)^23 - 1))/r - 366000 = 0
``````

Results plotted for r = 0.02, 0.025, 0.03, 0.035, 0.04

``````with  d = 11000
n = 23
s = 366000
``````

Zero error at `r = 0.032`

Or you can find a solver, e.g. https://www.wolframalpha.com

If you want a simple calculation, you have 253000 of principal, which gives the interest as 366000-253000 = 113000, which is 44.66% of the principal. That's an arithmetic average of 1.94% per year. We can make further adjustments to that. For instance, each dollar of principal is earning interest, on average, for only half the time period, so that gives 3.88%. Taking compounding into account gives a lower amount, but 3.88% is a good start for Newton's method. Obviously it's not good enough for an exam question, but for a lot of real-world needs, it's a good estimate.