Suppose that you want payments every year but instead of each payment being the same, you want to be some multiple of the previous payment. For example, you may want the payments to increase every year at $ 3\%$ to keep up with inflation. So we're going to introduce a growth term into the formula.

So imagine I want to receive $ \$100 $ per year for 10 years but I want that $ \$100 $ to grow at $ 3\% $ (in other words, I want $ \$100 $ in real terms, assuming $ 3\% $ inflation.) The stream of payments will look like:

$$ \$100.00, \$103.00, \$106.09, \$109.27, \$112.55, \$115.93, \$119.41, \$122.99, \$126.68, \$130.48$$

We're going to discount that at some rate $i$:

$$PV = \frac{\$100}{(1+i)} + \frac{\$100 \times (1.03)}{(1+i)^2} + \frac{\$100 \times (1.03)^2}{(1+i)^3} + . . . + \frac{\$100 \times (1.03)^9}{(1+i)^{10}}$$

So far so good. So let's generalize this a bit. Let $C$ be the payment amount, $g$ be the growth rate and $n$ be the number of payments. Then we get something that looks like this:

\[\begin{align*}

PV &= \frac{C}{1+i}+\frac{C (1+g)}{(1+i)^2}+. . . +\frac{C(1+g)^{n-1}}{(1+i)^n} \\

&= \frac{C}{1+i} + \frac{C}{1+i} \left(\frac{1+g}{1+i}\right)^1+\frac{C}{1+i} \left(\frac{1+g}{1+i}\right)^2+. . . +\frac{C}{1+i} \left(\frac{1+g}{1+i}\right)^{n-1} \\

&= \sum_{j=0}^{n-1}\frac{C}{1+i}\left(\frac{1+g}{1+i}\right)^j

\end{align*}\]

This looks just like our old friend the geometric series. Here our $a$ and $r$ are given by:

\[\begin{align*}

a &= \frac{C}{1+i} \\

r &= \frac{1+g}{1+i}

\end{align*}\]

So our geometric progression annuity will look like this:

\[\begin{align*}

PV &= a\frac{1-r^n}{1-r} \\

&= \frac{C}{1+i} \frac{1-\left( \frac{1+g}{1+i} \right)^n}{1-\frac{1+g}{1+i}} \\

&= C \frac{1-\left( \frac{1+g}{1+i} \right)^n}{i-g}

\end{align*}\]

And that gives us a general formula for annuity with a stream of payments that follow a geometric progression.

__Example__

So let's go back to our original example in which we were consider $C=\$100$ payments, a growth rate of $g=3\%$, $n=10$ total payments and let's assume that the effective interest rate is $i = 4\%$. Now you can either plug that into the first equation we gave at the top (which would be a lot of work in my opinion) or we can plug that into the equation we just derived at the bottom (I'm lazy; so that's my preference). So let's work it with the equation we just derived.

\[\begin{align*}

PV &= C \frac{1-\left( \frac{1+g}{1+i} \right)^n}{i-g} \\

&= \$100 \frac{1-\left( \frac{1+3\%}{1+4\%} \right)^{10}}{4\% - 3\%} \\

&= \$920.98

\end{align*}\]

Now can we actually calculate this using a financial calculator? The answer is "yes" but we'll need to fudge the formulas a bit.

With the geometric progression, we don't simply have an interest rate term; we have a growth and interest rate term. As a result, we need a new "interest rate" for the purpose of plugging it into a financial calculator. We'll call this new interest rate $i^*$ so as to not confuse it with the other interest rate.

Recall that we defined $v=(1+i)^{-1}$. So we're going to create a new $v$ and then solve for our new $i^*$.

For the geometric progression, our $v$ term is given by:

$$v = \frac{1+g}{1+i} = \frac{1}{1+i^*}$$

We will now solve for $i^*$.

\[\begin{align*}

\frac{1+g}{1+i} &= \frac{1}{1+i^*} \\

1+i^* &= \frac{1+i}{1+g} \\

i^* &= \frac{1+i}{1+g} -1

\end{align*}\]

So we're going to use this $i^*$ in the financial calculator to find the present value. For our example, $i^* = 0.97\%$.

We also need to adjust our payment term by the growth rate. So our new payment term will be $C^*=C/(1+g)$. For our example, our new payment term will give given by $C^*=\$97.09$.

This will enable us to use the financial calculators.

For Excel, we will need the following command:

=PV(0.97%, 10, 97.09) = ($921.05)

For the BAII Plus, after clearing we'll need to hit 10 "N", 0.97 "I/Y", 97.09 "PMT" and hit "CPT" "PV" to calculate -921.05.

You'll notice that the result is slightly different than the $920.98 we initially calculated. This is due to rounding. Had we inputted more precise numbers, we would have gotten the same result.

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