Accumulated Values of Geometric Annuities

Accumulating a Geometric Payment Stream

A geometric annuity has payments that change by the same percentage from one payment to the next. When the question asks for the amount at the final payment date, each earlier payment must earn interest from its own payment date to that final date. The resulting accumulated value is not the present value: it is measured at a later point in time.

Assume payments occur at the end of each period, the first payment is $P$ at the end of period 1, and each subsequent payment is the preceding payment multiplied by $1+g$. Here $g$ is the effective growth rate per payment period. Let $i$ be the effective interest rate per period, and suppose it is constant throughout the term. The final payment is at the end of period $n$.

Definition: The accumulated value at time $n$ is the sum of every payment accumulated from its payment date to time $n$. For payments $P(1+g)^{k-1}$ at the ends of periods $k=1,\ldots,n$, with a constant effective rate $i$, it is $$ A_n=\sum_{k=1}^{n}P(1+g)^{k-1}(1+i)^{n-k}. $$

The exponent $n-k$ counts how many periods the payment at time $k$ earns interest. The first payment earns interest for $n-1$ periods; the last payment, made at time $n$, earns no interest before the valuation date. This timing convention is essential to the formula.

The Accumulated-Value Formula

For $i\ne g$, the sum has a closed form. This expression is useful when the interest and payment-growth rates differ:

A_n=P\frac{(1+i)^n-(1+g)^n}{i-g},\qquad i\ne g.

If payments decrease by a fixed percentage $h$ each period, write $g=-h$. Provided $1-h>0$, the payment at time $k$ is $P(1-h)^{k-1}$, and the formula becomes

A_n=P\frac{(1+i)^n-(1-h)^n}{i+h},\qquad i\ne -h.

The formula assumes a level interest rate per period and a geometric sequence of payments at consecutive period ends. If a problem gives a nominal annual rate convertible more than once per year while payments occur on a different schedule, first establish the effective rate per payment period. Do not apply this formula unchanged to beginning-of-period payments, deferred payments, or irregular payment dates.

Why the Formula Holds

The formula follows directly from accumulating the individual payments. Factoring out the interest accumulation that the first payment would receive gives a finite geometric series:

\begin{aligned} A_n &=P\sum_{k=1}^{n}(1+g)^{k-1}(1+i)^{n-k}\\ &=P(1+i)^{n-1}\sum_{j=0}^{n-1} \left(\frac{1+g}{1+i}\right)^j. \end{aligned}

For a finite geometric series, $\sum_{j=0}^{n-1}r^j=(1-r^n)/(1-r)$ when $r\ne1$. Substituting $r=(1+g)/(1+i)$, which is not 1 when $i\ne g$, gives

\begin{aligned} A_n &=P(1+i)^{n-1} \frac{1-\left(\frac{1+g}{1+i}\right)^n} {1-\frac{1+g}{1+i}}\\ &=P\frac{(1+i)^n-(1+g)^n}{i-g}. \end{aligned}

When $i=g$, the ratio $r$ equals 1, so the finite-series formula with denominator $1-r$ cannot be used. Return to the payment-by-payment sum instead. Each payment accumulates to the same amount at time $n$, because payment growth exactly offsets the difference in accumulation time:

P(1+i)^{k-1}(1+i)^{n-k} =P(1+i)^{n-1}.

There are $n$ payments, so the special case is

Special case: If $i=g$, then $A_n=nP(1+i)^{n-1}$. Do not substitute $i=g$ into the formula with denominator $i-g$; that produces division by zero.

Worked Example: An Increasing Stream Accumulated to Its Final Date

A fund receives a payment of $1,000 at the end of year 1. Each annual payment is 5% greater than the preceding one, and the fund earns an effective annual interest rate of 10%. Find the accumulated value immediately after the payment at the end of year 4. Assume the stated rates apply for all four years.

Method. Use the geometric-annuity accumulated-value formula, with $P=1{,}000$, $g=0.05$, $i=0.10$, and $n=4$. Since $i\ne g$, the closed form applies.

The four payments are $1,000, $1,050, $1,102.50, and $1,157.625. The final payment is included at time 4 and does not earn interest before that date. Applying the formula:

\begin{aligned} A_4 &=1{,}000\frac{(1.10)^4-(1.05)^4}{0.10-0.05}\\ &=1{,}000\frac{1.4641-1.21550625}{0.05}\\ &=1{,}000\frac{0.24859375}{0.05}\\ &=1{,}000(4.971875)\\ &=\$4{,}971.875. \end{aligned}

Check by accumulating each payment separately: $1,000 earns three years of interest and becomes $1,331; $1,050 earns two years and becomes $1,270.50; $1,102.50 earns one year and becomes $1,212.75; and $1,157.625 earns no interest. The sum is $1{,}331+1{,}270.50+1{,}212.75+1{,}157.625=\$4{,}971.875$, matching the formula. Rounded to the nearest cent, the answer is $4,971.88.

Common mistake avoided: Treating all four payments as though each earned four years of interest would value them at the wrong date. Their accumulation periods are respectively three, two, one, and zero years.

Worked Example: Growth Rate Equal to the Interest Rate

A scholarship account receives $800 at each year-end for five years. Each payment is 4% greater than the previous payment, and the account earns an effective annual interest rate of 4%. Find its accumulated value immediately after the fifth payment.

Method. Here $P=800$, $g=i=0.04$, and $n=5$. Because the rates are equal, use the special case rather than the formula with denominator $i-g$.

\begin{aligned} A_5 &=5(800)(1.04)^{5-1}\\ &=4{,}000(1.16985856)\\ &=\$4{,}679.43424. \end{aligned}

For a check, the payments are $800, $832, $865.28, $899.8912, and $935.886848. At time 5, each has accumulated to $800(1.04)^4=\$935.886848$: for example, the second payment grows for three years, so $832(1.04)^3=\$935.886848$. Thus the five accumulated payments total $5(935.886848)=\$4{,}679.43424$. Rounded to the nearest cent, the accumulated value is $4,679.43.

Common mistake avoided: Substituting equal rates into the unequal-rate formula gives a zero denominator. The direct sum, or its special-case result, is required.

Decreasing Payments and Present-Value Conversion

A percentage-decreasing sequence is handled by making its growth rate negative. For a decrease of $h$ per period, the payments are $P, P(1-h), P(1-h)^2,\ldots$. The accumulated value can be found either by accumulating those payments directly or by converting a present value to time $n$. Under the same constant-rate and end-of-period assumptions, multiplying a time-0 value by $(1+i)^n$ moves that value to time $n$.

For example, the present value of this payment stream is $P\sum_{k=1}^n(1-h)^{k-1}(1+i)^{-k}$. Multiplying by $(1+i)^n$ changes each term to $P(1-h)^{k-1}(1+i)^{n-k}$, exactly the amount of that payment accumulated to time $n$. Thus the present-value conversion and the direct accumulation method agree.

Worked Example: A Percentage-Decreasing Payment Stream

A project receives $1,200 at the end of year 1. Each subsequent annual payment is 10% less than the preceding payment. The project earns an effective annual interest rate of 5%. Find the accumulated value immediately after the payment at the end of year 6.

Method. Set $P=1{,}200$, $h=0.10$, $i=0.05$, and $n=6$. The payment-growth rate is $g=-0.10$, so $i-g=0.15$. Apply the decreasing-payment form:

\begin{aligned} A_6 &=1{,}200\frac{(1.05)^6-(0.90)^6}{0.05+0.10}\\ &=1{,}200\frac{1.340095640625-0.531441}{0.15}\\ &=1{,}200\frac{0.808654640625}{0.15}\\ &=1{,}200(5.3910309375)\\ &=\$6{,}469.237125. \end{aligned}

Check the timing by listing the accumulated payments: $1{,}200(1.05)^5=\$1{,}531.537875$; $1{,}080(1.05)^4=\$1{,}312.74675$; $972(1.05)^3=\$1{,}125.2115$; $874.80(1.05)^2=\$964.467$; $787.32(1.05)=\$826.686$; and the final payment is $708.588$. Their sum is $6,469.237125, agreeing with the formula. Rounded to the nearest cent, the accumulated value is $6,469.24.

Common mistake avoided: A 10% decrease means multiplying each payment by $0.90$, not by $0.10$. The latter would make each payment only one-tenth of the previous amount.

Common Timing and Rate Traps

The formula is compact, but a correct calculation still depends on matching the payment timing, rate period, and valuation date. Check each of these before substituting numbers.

Use the rate per payment period. If payments are annual, $i$ and $g$ in the formula must both describe annual changes. Convert rates appropriately if the problem specifies a different compounding or payment frequency.
Count accumulation periods from each payment date. For end-of-period payments valued at time $n$, payment $k$ earns interest for $n-k$ periods. In particular, the last payment earns none.
Keep the valuation date distinct from time 0. A present value is measured at time 0. To obtain the accumulated value at time $n$, accumulate it for $n$ periods under the same interest assumptions.
Check the equal-rate case before dividing. If $i=g$, use $nP(1+i)^{n-1}$. If rates are close but unequal, the formula is valid, although careful arithmetic may be needed.
Interpret decreases as negative growth. A decrease of $h$ corresponds to $g=-h$, so the denominator is $i+h$, not $i-h$.

Key takeaway: For end-of-period geometric payments, accumulate payment $k$ for $n-k$ periods. When $i\ne g$, the accumulated value is $P\big((1+i)^n-(1+g)^n\big)/(i-g)$; when $i=g$, it is $nP(1+i)^{n-1}$.

Check Your Understanding

For each problem, assume payments occur at consecutive year-ends and the stated effective annual interest rate remains constant. Round money to the nearest cent.

An account receives $500 at the end of year 1, and each payment grows by 2% annually. If the effective annual interest rate is 6%, find the accumulated value immediately after the third payment. (Answer: $1,622.60)
A fund receives $2,000 at the end of year 1, and each following payment is 5% smaller than the preceding one. At an effective annual interest rate of 8%, find the accumulated value at the end of year 4. (Answer: $8,399.73)
A program receives $900 at the end of year 1, with each payment growing by 3.5% annually. The account earns 3.5% effective annual interest. Find the accumulated value immediately after payment 6. (Answer: $6,413.50)
A trust receives $750 at the end of year 1, and each following payment grows by 4%. If its effective annual interest rate is 7%, calculate the accumulated value at the end of year 5. (Answer: $4,647.47)
An account receives $1,000 at the end of year 1, and each following payment grows by 6%. The effective annual interest rate is 4%. Find the accumulated value immediately after payment 7. (Answer: $9,384.92)

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