Reinvesting Arithmetic and Geometric Cash Flows

Why the Pattern of Payments Matters

When payments change over time, their terminal value depends both on the amount of each payment and on how long it is reinvested. A sequence that increases by a fixed dollar amount is arithmetic; one that increases by a fixed percentage is geometric. Recognizing which pattern applies can turn a sum of many accumulated payments into a compact formula, but only when the cash-flow timing and reinvestment assumptions match the formula.

Throughout this tutorial, assume payments occur at the ends of periods 1 through $n$, and the terminal date is the end of period $n$. Each payment is reinvested immediately at the same effective rate $i$ per period, with no fees, taxes, defaults, or withdrawals. Thus a payment made at the end of period $k$ earns interest for $n-k$ periods. These formulas do not apply unchanged if payment dates or reinvestment rates differ.

Definition: The terminal value at time $n$ is the sum of all payments accumulated to time $n$. Under a constant effective rate $i$, a payment $C_k$ at the end of period $k$ contributes $C_k(1+i)^{n-k}$, so the terminal value is $\sum_{k=1}^{n} C_k(1+i)^{n-k}$.

Write $q=1+i$ for the accumulation factor per period. For $i\ne 0$, the accumulated value of $n$ level end-of-period payments of 1 is $s_{\overline{n}|}=(q^n-1)/i$. This is the level-payment result; changing payments can be handled by applying it to a level component or by summing the accumulated payments directly. If $i=0$, every payment remains unchanged, so the terminal value is simply the sum of the payments.

Arithmetic Changes: A Fixed Dollar Increase

Suppose the first payment is $A$, and each later payment is $G$ dollars larger than the previous one. The payment at the end of period $k$ is $A+(k-1)G$. The increase $G$ is an amount, not a percentage: for example, payments of 100, 120, 140, and 160 have a constant increase of 20.

\text{Terminal value} =\sum_{k=1}^{n}\bigl(A+(k-1)G\bigr)q^{n-k} =A\,s_{\overline{n}|} +G\frac{q^n-nq+n-1}{i^2},\qquad i\ne 0.

The second term is the accumulated value of the increase pattern $0,G,2G,\ldots,(n-1)G$. When $i=0$, the same payments total $nA+G\,n(n-1)/2$. The formula with $i^2$ in the denominator is not used at $i=0$; use the cash-flow sum or this zero-rate result instead.

Justification: Set $j=n-k$, the number of periods for which a payment is reinvested. Then $k-1=n-1-j$, so the unit increase component is $\sum_{j=0}^{n-1}(n-1-j)q^j$. Using $\sum_{j=0}^{n-1}q^j=(q^n-1)/(q-1)$ and $\sum_{j=0}^{n-1}jq^j=[q-nq^n+(n-1)q^{n+1}]/(q-1)^2$, subtracting the second sum from $n-1$ times the first gives $(q^n-nq+n-1)/(q-1)^2$. Since $q-1=i$, this is the stated factor. The derivation assumes a constant rate and end-of-period payments.

Worked Example: Arithmetic Payments Reinvested at a Constant Rate

A hypothetical reserve receives 100 dollars at the end of year 1, with each annual payment increasing by 20 dollars. There are four payments, and each is reinvested at an effective annual rate of 5% until the end of year 4. Find the terminal value. Use the payment-by-payment method and verify it with the arithmetic formula.

Method and setup. Here $A=100$, $G=20$, $n=4$, $i=0.05$, and $q=1.05$. The payments are 100, 120, 140, and 160. They earn interest for 3, 2, 1, and 0 years, respectively.

Accumulate each payment. The four contributions at year 4 are $100(1.05)^3=115.7625$, $120(1.05)^2=132.3000$, $140(1.05)=147.0000$, and $160$. Their sum is $115.7625+132.3000+147.0000+160=555.0625$ dollars.

Check with the formula. The level component is $100s_{\overline{4}|}=100[(1.05)^4-1]/0.05=100(4.310125)=431.0125$. For the increase component, $[(1.05)^4-4(1.05)+3]/(0.05)^2=[1.21550625-4.2+3]/0.0025=6.2025$. Multiplying by $G=20$ gives $124.0500$. Thus the formula gives $431.0125+124.0500=555.0625$ dollars, matching the direct sum.

Answer and trap avoided. The terminal value is $555.06, rounded to the nearest cent. The common mistake is to treat every payment as if it earns three years of interest. The last payment is made at the terminal date and earns no interest; the exponents must run from 3 down to 0.

Geometric Changes: A Fixed Percentage Increase

If each payment is a fixed multiple of the previous payment, the sequence is geometric. Let the first payment be $P$, and let $r$ be the ratio between consecutive payments. The payment at the end of period $k$ is $Pr^{k-1}$. For an annual increase of $g$, the ratio is $r=1+g$, where $g$ is expressed as a decimal. This is different from an arithmetic increase: a 5% increase is applied to each successive payment, so the dollar increase need not stay constant.

\text{Terminal value} =P\sum_{k=1}^{n}r^{k-1}q^{n-k} = \begin{cases} P\dfrac{q^n-r^n}{q-r},&q\ne r,\\[6pt] Pnq^{n-1},&q=r. \end{cases}

For $q\ne r$, the accumulated terms form a finite geometric sum: the first is $q^{n-1}$, the last is $r^{n-1}$, and the common ratio from one term to the next is $r/q$. Equivalently, summing $q^{n-1}+rq^{n-2}+\cdots+r^{n-1}$ gives $(q^n-r^n)/(q-r)$. If $q=r$, every payment accumulates to the same amount $Pq^{n-1}$, so the total is $nPq^{n-1}$. The first formula cannot be used when its denominator is zero.

Worked Example: Geometric Payments With a Different Growth Rate

A hypothetical fund receives four annual payments. The first is 80 dollars, and each later payment is 25% larger than the preceding payment. All payments are reinvested at an effective annual rate of 4% until the end of year 4. Calculate the terminal value.

Method and setup. The payment ratio is $r=1.25$, while the accumulation factor is $q=1.04$. Since $q\ne r$, use the geometric terminal-value formula. The payments are $80$, $80(1.25)=100$, $100(1.25)=125$, and $125(1.25)=156.25$.

Accumulate each payment. The contributions at year 4 are $80(1.04)^3=89.98912$, $100(1.04)^2=108.16000$, $125(1.04)=130.00000$, and $156.25$. Their sum is $89.98912+108.16000+130.00000+156.25=484.39912$ dollars.

Check with the formula. The powers are $q^4=(1.04)^4=1.16985856$ and $r^4=(1.25)^4=2.44140625$. Therefore, $80(q^4-r^4)/(q-r)=80(1.16985856-2.44140625)/(1.04-1.25)=80(-1.27154769)/(-0.21)=484.39912$ dollars.

Answer and trap avoided. The terminal value is $484.40, rounded to the nearest cent. The negative numerator and negative denominator are both correct and produce a positive value. Reversing only one subtraction would incorrectly produce a negative terminal value.

Worked Example: Payment Growth Equals the Reinvestment Rate

A hypothetical account receives a first payment of 50 dollars at the end of year 1. Each later payment is 10% larger than the previous one. There are three payments, and the effective annual reinvestment rate is also 10%. Find the terminal value at the end of year 3.

Method and setup. Here $P=50$, $n=3$, $r=1.10$, and $q=1+i=1.10$. Because $r=q$, the formula with $q-r$ in the denominator is undefined, so use the equal-ratio case.

Accumulate and check. The payments are $50$, $55$, and $60.50$. At year 3, their accumulated values are $50(1.10)^2=60.50$, $55(1.10)=60.50$, and $60.50$. The sum is $181.50$ dollars. The special-case formula agrees: $Pnq^{n-1}=50(3)(1.10)^2=181.50$.

Answer and trap avoided. The terminal value is $181.50. The equal-ratio condition makes each payment grow at exactly the rate needed to offset its shorter reinvestment period, so every accumulated contribution is equal. Do not substitute $q=r$ into a quotient whose denominator is zero.

Selecting and Checking a Formula

Before using a closed form, identify the payment pattern from the amounts themselves. A fixed difference indicates an arithmetic sequence; a fixed ratio indicates a geometric sequence. The accumulation factor $q$ is determined separately by the reinvestment rate. A cash-flow sequence can therefore be geometric even when its growth ratio differs from the investment accumulation factor.

Payment pattern	Payment at time $k$	Terminal-value approach
Level	$A$	$A s_{\overline{n}\|}$
Arithmetic	$A+(k-1)G$	Level component plus the fixed-dollar increase component
Geometric	$Pr^{k-1}$	Finite geometric sum; use the special case if $r=q$

A reliable check is to write at least the first and last accumulated contributions. For end-of-period payments, the first payment has $n-1$ accumulation periods and the last has none. Then check that the formula's inputs reflect the stated number of payments, not the number of intervals between them. For example, four payments at times 1, 2, 3, and 4 have exponents 3, 2, 1, and 0.

Describe the payments.
Write the payment at period $k$. Use $A+(k-1)G$ for a fixed dollar increase or $Pr^{k-1}$ for a fixed percentage increase.

Set the terminal date and rate.
For payments at times 1 through $n$ accumulated to time $n$, use $q=1+i$ and the exponent $n-k$ for payment $k$.

Choose and check the sum.
Use the arithmetic formula for a constant dollar difference, the geometric formula for a constant ratio, and the equal-ratio case when $r=q$. Compare the result with the individual-payment sum.

The main exam trap is mixing the payment growth pattern with the accumulation pattern. Arithmetic payments do not become geometric merely because they earn compound interest, and a geometric payment ratio does not replace the reinvestment factor. Each payment must still be accumulated using the number of periods it actually earns interest. These methods are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, assume end-of-period payments and a constant effective annual reinvestment rate through the stated terminal date.

A fund receives 200 dollars at the end of year 1, followed by payments increasing by 25 dollars each year, for three payments total. The effective annual rate is 5%. Find the terminal value at year 3. (Answer: $706.75)
A first payment of 120 dollars is followed by payments that are 3% larger each year. There are three payments, the effective annual rate is 2%, and the terminal date is year 3. Find the terminal value. (Answer: $378.23)
Four payments are 75, 90, 105, and 120 dollars, respectively. They are reinvested at an effective annual rate of 6% through year 4. Find the terminal value. (Answer: $421.75)
A first payment of 64 dollars grows by 5% per year for a total of four payments. The effective annual reinvestment rate is also 5%, and the terminal date is year 4. Find the terminal value. (Answer: $296.35)
A first payment of 90 dollars grows by 3% per year for five payments. The effective annual reinvestment rate is 4%, and the terminal date is year 5. Use the geometric formula to find the terminal value. (Answer: $516.41)

Payment pattern	Payment at time \(k\)	Terminal-value approach
Level	\(A\)	\(A s_{\overline{n}\|}\)
Arithmetic	\(A+(k-1)G\)	Level component plus the fixed-dollar increase component
Geometric	\(Pr^{k-1}\)	Finite geometric sum; use the special case if \(r=q\)

What You'll Learn