Accumulated Values of Arithmetic-Gradient Annuities

Why Accumulated Values Need Careful Timing

An arithmetic-gradient annuity has payments that change by a fixed dollar amount from one payment date to the next. To find its accumulated value, the payment amounts alone are not enough: each payment must also be carried forward for the correct number of periods. For an annuity-immediate, payments are made at the ends of periods, and the accumulated value is measured immediately after the final payment. The last payment earns no interest before that valuation date.

This tutorial develops accumulated-value formulas for both increasing and decreasing arithmetic-gradient annuities. The same result can be obtained by accumulating each payment separately or by finding present value and then accumulating that value to the final payment date.

Definition: An arithmetic-increasing annuity-immediate with $n$ payments has payment $A+(k-1)G$ at time $k$, for $k=1,\ldots,n$, where $A$ is the first payment and $G$ is the fixed increase per period. An arithmetic-decreasing annuity has payment $A-(k-1)D$, where $D$ is the fixed decrease per period. Payments occur at the ends of periods. The accumulated value is measured at time $n$, immediately after payment $n$.

The Accumulation Formula

Assume a constant effective interest rate $i$ per payment period, with $i\ne 0$. A payment made at time $k$ accumulates for $n-k$ periods, so it is multiplied by $(1+i)^{n-k}$. For an increasing annuity, this gives

AV_n=\sum_{k=1}^{n}\bigl(A+(k-1)G\bigr)(1+i)^{n-k}.

Separate the level-payment part from the gradient part. The level payments accumulate to $A s_{\overline{n}|}$. The gradient payments have amounts $0,G,2G,\ldots,(n-1)G$, so their accumulated value is $G$ times the accumulation factor for the sequence $0,1,2,\ldots,n-1$:

AV_n=A s_{\overline{n}|} +G\sum_{k=1}^{n}(k-1)(1+i)^{n-k} =A s_{\overline{n}|} +G\frac{(1+i)^n-1-ni}{i^2}.

The first payment has gradient component zero, the second has gradient component $G$, and the final payment has gradient component $(n-1)G$. That indexing is essential: a sequence $G,2G,\ldots,nG$ is a different payment pattern.

Theorem: For $n$ end-of-period payments $A+(k-1)G$, under a constant effective rate $i$ per period, the accumulated value immediately after the $n$th payment is $$ AV_n=A s_{\overline{n}|} +G\frac{(1+i)^n-1-ni}{i^2}. $$ For decreasing payments $A-(k-1)D$, replace $G$ by $-D$. If $i=0$, the accumulated value is the sum of the payments: $nA+G n(n-1)/2$ for the increasing case, or $nA-D n(n-1)/2$ for the decreasing case.

Why the Gradient Factor Has This Form

Let $D_n$ denote the accumulated value of the unit gradient sequence $0,1,2,\ldots,n-1$ at time $n$, at rate $i$. Its direct expression is

D_n=\sum_{k=1}^{n}(k-1)(1+i)^{n-k}.

When one more period and payment are added, the previous accumulated value earns one period of interest, and the new gradient payment is $n$. Therefore $D_{n+1}=(1+i)D_n+n$, with $D_1=0$. The claimed expression follows by induction. If $D_n=((1+i)^n-1-ni)/i^2$, then

\begin{aligned} D_{n+1} &=(1+i)\frac{(1+i)^n-1-ni}{i^2}+n\\ &=\frac{(1+i)^{n+1}-1-(n+1)i}{i^2}. \end{aligned}

The base case $n=1$ also agrees, because $((1+i)-1-i)/i^2=0$. Thus the formula holds for every positive integer $n$. This recurrence also explains why the final payment’s gradient component is $n$ when moving from $n$ payments to $n+1$: it is the new payment, not an extra period of interest on the previous final payment.

Two Routes to the Accumulated Value

The direct route is usually safest when the number of payments is small: write each payment, count how many periods it earns interest, and add. The closed-form factor is more efficient for larger $n$, and it makes the separate contributions of the level amount and gradient clear.

There is also a present-value route. If $PV$ is the value at time 0 of the same payment stream, then each payment is accumulated from time 0 to time $n$ by the same factor $(1+i)^n$. Hence $AV_n=(1+i)^nPV$. For an increasing annuity, the present value can be written as $A a_{\overline{n}|}+G(Ia)_{\overline{n}|}$, where $(Ia)_{\overline{n}|}$ is the present value of payments $1,2,\ldots,n$ at times $1,2,\ldots,n$. Accumulating this present value gives the same result as the direct formula. This conversion assumes the same effective rate applies over all periods.

Set the payment pattern.
Write the payment at time $k$, such as $A+(k-1)G$ or $A-(k-1)D$. Check whether the first payment is $A$, not $A+G$.

Set the valuation date.
For an annuity-immediate accumulated just after payment $n$, payment $k$ earns interest for exactly $n-k$ periods.

Calculate by a chosen route.
Use the payment-by-payment sum, the closed-form accumulated-value factor, or present value multiplied by $(1+i)^n$.

Check the result.
Confirm that the final payment earns no interest, the gradient has the correct sign, and the accumulated value is consistent with the payment amounts.

Worked Applications

Worked Example: Accumulating an Increasing Payment Stream

A fund receives three deposits at the ends of years 1, 2, and 3. The deposits are $100, $120, and $140, respectively. The effective annual interest rate is 10%. Find the fund value immediately after the third deposit.

Method. This is an annuity-immediate with $A=\$100$, $G=\$20$, $n=3$, and $i=0.10$. Each deposit is carried to time 3 for the number of years remaining.

The first deposit earns interest for two years, the second for one year, and the third for zero years:

\begin{aligned} AV_3 &=100(1.10)^2+120(1.10)+140\\ &=100(1.21)+132+140\\ &=121+132+140\\ &=\$393.00. \end{aligned}

Using the formula gives the same result. Here $s_{\overline{3}|}=(1.10^3-1)/0.10=3.31$, and the gradient factor is $(1.10^3-1-3(0.10))/(0.10)^2=(1.331-1.3)/0.01=3.1$. Thus $100(3.31)+20(3.1)=331+62=\$393.00$. The direct sum confirms the formula. A common mistake would be to give the third deposit one year of interest; it is made at the valuation date and earns none.

Worked Example: Finding the First Deposit From a Target Value

Four year-end deposits increase by $25 each year. The effective annual interest rate is 5%, and the accumulated value immediately after the fourth deposit is $600. Find the first deposit, to the nearest cent.

Method. Let the first deposit be $A$. The deposits are $A,A+25,A+50,A+75$. The first three deposits earn three, two, and one years of interest; the last earns none.

\begin{aligned} 600 &=A(1.05)^3+(A+25)(1.05)^2+(A+50)(1.05)+(A+75)\\ &=A(1.157625)+A(1.1025)+A(1.05)+A\\ &\quad+25(1.1025)+50(1.05)+75\\ &=4.310125A+27.5625+52.5+75\\ &=4.310125A+155.0625. \end{aligned}

Subtract the accumulated gradient component and divide by the accumulated-value factor for the four level payments:

A=\frac{600-155.0625}{4.310125} =\frac{444.9375}{4.310125} \approx 103.2307648.

Rounded to the nearest cent, the first deposit is $103.23. As a check, substituting $103.23 gives $4.310125(103.23)+155.0625=\$599.99670375$, which rounds to $600.00. The small difference from the target is due to rounding the deposit to cents. A common error is to miscalculate the gradient component or the denominator; the displayed decomposition keeps the level and increasing parts separate.

Worked Example: Accumulating a Decreasing Annuity

A hypothetical reserve receives five deposits at the ends of successive years. The first deposit is $200, and each later deposit is $30 less than the preceding one. The effective annual interest rate is 4%. Find the accumulated value immediately after the fifth deposit.

Method. The payment sequence is $200, $170, $140, $110, and $80. Treat the decreases as a negative gradient: $A=200$, $G=-30$, $n=5$, and $i=0.04$.

The level-payment accumulation factor is $s_{\overline{5}|}=((1.04)^5-1)/0.04=(1.2166529024-1)/0.04=5.41632256$. The unit gradient factor is $(1.2166529024-1-5(0.04))/(0.04)^2=0.0166529024/0.0016=10.408064$. Therefore,

\begin{aligned} AV_5 &=200(5.41632256)-30(10.408064)\\ &=1083.264512-312.24192\\ &=\$771.022592. \end{aligned}

The accumulated value is $771.02 to the nearest cent. As a direct check, accumulating the five payments gives $200(1.04)^4+170(1.04)^3+140(1.04)^2+110(1.04)+80$ $=233.971712+191.22688+151.424+114.4+80=\$771.022592$. The direct calculation agrees. The decreasing payments are still positive, so the negative gradient reduces the value relative to five level deposits of $200; it does not make the accumulated value negative.

Common Errors and Practical Checks

Accumulated values are sensitive to timing and indexing. Before using a factor, specify the payment dates, the rate per payment period, and the precise valuation date. If payments are annual, $i$ must be the effective annual rate; if periods are monthly, the rate used must be an effective rate per month. A nominal annual rate cannot be inserted as though it were an effective rate per payment period without first converting it appropriately.

Giving the last payment interest. A value immediately after the final annuity-immediate payment includes that payment at face value, not multiplied by $1+i$.
Starting the gradient at the wrong amount. In this convention the payment sequence is $A,A+G,A+2G,\ldots$, so the gradient component of the first payment is zero.
Using an increasing-gradient sign for a decreasing stream. A decrease of $D$ per period means the gradient is $-D$, or equivalently the accumulated gradient factor is subtracted.
Mixing up periods and years. If the payment interval is one year, a rate and exponent measured in months do not match the payment schedule.
Rounding too early. Keep the accumulation factors and intermediate values unrounded until the final money amount.

The formula presumes a fixed rate per period and regular end-of-period payments. It does not apply unchanged when the interest rate varies by period, payment dates are irregular, or payments occur at the beginning of periods. Those cases require matching each payment to its actual accumulation interval.

Key takeaway: For an arithmetic-gradient annuity-immediate, carry each payment to the date immediately after the last payment. The level part accumulates by $s_{\overline{n}|}$, and the gradient part by $\bigl((1+i)^n-1-ni\bigr)/i^2$; use a negative gradient for decreasing payments.

Check Your Understanding

For each problem, assume a constant effective annual interest rate and annual end-of-year payments. Accumulated values are measured immediately after the final payment.

A fund receives deposits of $80, $100, and $120 at the ends of years 1, 2, and 3. The effective annual rate is 6%. Find the accumulated value. (Answer: $315.89)
Four annual deposits are $150, $140, $130, and $120. At an effective annual rate of 5%, find the accumulated value immediately after deposit 4. (Answer: $584.49)
Three deposits are $A$, $A+\$15$, and $A+\$30$. At an effective annual rate of 4%, the accumulated value after deposit 3 is $500. Find $A$ to the nearest cent. (Answer: $145.57)
Six annual deposits begin at $180 and decrease by $20 each year. At an effective annual rate of 3%, find the accumulated value after the sixth deposit, to the nearest cent. (Answer: $852.04)
Two annual deposits are $100 and $120, and the effective annual rate is 10%. Find the accumulated value just after the second deposit. Then state how much of that value is interest earned by the first deposit. (Answer: accumulated value $230.00; interest earned by the first deposit $10.00)

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