Geometric-Decreasing Annuities and Terminal Payments

Introduction

A declining payment stream and a separate terminal payment can have very different timing, even when both happen near the end of a contract. The reliable approach is to put every payment on a timeline, identify the date of the last regular installment, and then value each cash flow at one common focal date. This is especially important when the stream is deferred: the first payment is not necessarily one period from now.

This tutorial considers a finite sequence of payments that declines by a constant proportion from one payment to the next, together with an additional lump sum. The lump sum may be paid on the same date as the final regular installment or on another specified date. All examples use annual payments and a constant effective annual interest rate. No mortality or other contingent-payment assumption is involved.

Definition: A geometric-decreasing annuity has $n$ regular payments $C_1,C_2,\ldots,C_n$, with $C_1>0$ and $C_k=C_1r^{k-1}$ for $k=1,\ldots,n$, where $0<r<1$. The number $r$ is the payment ratio: each regular payment is $r$ times the preceding payment. It is distinct from the interest rate.

Set the Timeline Before Using a Formula

Let $i$ be the effective annual interest rate and $v=1/(1+i)$ the annual discount factor. Suppose the stream is deferred for $m$ complete years, where $m$ is a nonnegative integer. Under the convention used here, the first regular payment occurs at time $m+1$, and the $n$th, or final, regular payment occurs at time $m+n$. Thus, a stream with $m=0$ is an annuity-immediate: its first payment is at time 1, not time 0.

If a separate terminal payment $L$ is made at time $T$, its present value is $Lv^T$. When it is paid at the same time as the final regular installment, $T=m+n$, and the two cash flows are added. The terminal payment does not replace the $n$th regular payment unless the problem explicitly says so.

Present-value result: For $n$ regular payments at times $m+1,\ldots,m+n$, followed by or accompanied by a separate payment $L$ at time $T$, the value at time 0 is $$ PV_0=C_1v^{m+1}\frac{1-(rv)^n}{1-rv}+Lv^T, $$ provided $rv\ne 1$. If $rv=1$, the regular-payment part is $nC_1v^{m+1}$. The formula assumes a constant effective annual rate, annual payment periods, and the stated payment dates.

The finite geometric sum in this result is a compact way to account for every installment. It does not remove the need to check the payment dates or determine whether the lump sum shares the last installment's date.

Why the Geometric-Value Formula Works

Each regular payment $C_k=C_1r^{k-1}$ is discounted from its actual payment date $m+k$. Its value at time 0 is therefore $C_1r^{k-1}v^{m+k}$. Factoring out the common first-period terms gives:

\begin{aligned} PV_{\text{regular}} &=\sum_{k=1}^{n} C_1r^{k-1}v^{m+k}\\ &=C_1v^{m+1}\sum_{j=0}^{n-1}(rv)^j\\ &=C_1v^{m+1}\frac{1-(rv)^n}{1-rv}. \end{aligned}

The last equality follows by multiplying the finite sum $1+x+\cdots+x^{n-1}$ by $1-x$, where $x=rv$: all intermediate terms cancel, leaving $1-x^n$. If $x=1$, the sum has $n$ terms each equal to 1, so its value is $n$. The separate payment is not part of this geometric sum unless it is explicitly one of the regular installments; its value is added as $Lv^T$.

Worked Examples

Worked Example: Immediate Declining Payments and a Same-Date Terminal Payment

A fund will pay four annual installments. The first is $1,000, and each subsequent installment is 50% of the preceding one. An additional $300 is paid on the date of the fourth installment. The effective annual interest rate is 25%. Find the value at time 0 and check it by accumulating all payments to time 4.

Method. Set $C_1=1000$, $r=0.5$, $n=4$, $m=0$, $i=0.25$, and $v=1/1.25=0.8$. The regular payments occur at times 1, 2, 3, and 4. The terminal payment is also at time 4, so it is additional to the fourth installment.

The regular payments are $1000$, $500$, $250$, and $125$. Discount each from its own payment date:

\begin{aligned} PV_{\text{regular}} &=1000(0.8)+500(0.8)^2+250(0.8)^3+125(0.8)^4\\ &=800+320+128+51.20\\ &=1299.20. \end{aligned}

The additional terminal payment has value $300(0.8)^4=300(0.4096)=122.88$. Therefore,

PV_0=1299.20+122.88=\$1,422.08.

Check at time 4. Accumulate each cash flow from its payment date to time 4:

\begin{aligned} AV_4 &=1000(1.25)^3+500(1.25)^2+250(1.25)+125+300\\ &=1953.125+781.25+312.50+125+300\\ &=\$3,471.875. \end{aligned}

Since $1.25^4=2.44140625$, $PV_0(1.25)^4=1422.08(2.44140625)=3471.875$, agreeing with the accumulated value. The answer is $1,422.08 at time 0, or $3,471.88 at time 4 when rounded to the nearest cent. The method avoids the common mistake of treating the terminal payment as a replacement for the final regular installment.

Worked Example: A Deferred Stream With a Terminal Payment

A contract pays three annual installments, beginning three years from now. The first installment is $1,200, and each later installment is half the preceding one. A separate $400 terminal payment is made on the date of the third installment. At an effective annual rate of 25%, find the value at time 0 and verify it at the terminal date.

Method. “Beginning three years from now” means the first payment is at time 3. Under the convention that the first payment is at $m+1$, this is $m=2$, not $m=3$. Here $n=3$, $C_1=1200$, $r=0.5$, and $v=0.8$. The regular payments occur at times 3, 4, and 5; the terminal payment is at time 5.

\begin{aligned} PV_{\text{regular}} &=1200(0.8)^3+600(0.8)^4+300(0.8)^5\\ &=614.40+245.76+98.304\\ &=958.464. \end{aligned}

The separate terminal payment has present value $400(0.8)^5=400(0.32768)=131.072$. Adding it gives:

PV_0=958.464+131.072=\$1,089.536.

To check, accumulate to time 5. The regular installments accumulate to $1200(1.25)^2+600(1.25)+300=1875+750+300=2925$. Adding the $400 terminal payment gives $AV_5=3325$. Discounting this value five years gives $3325(0.8)^5=3325(0.32768)=1089.536$, as required. The present value is $1,089.54 to the nearest cent. The timing check avoids discounting the first payment for four years merely because the stream has a two-year deferral.

Worked Example: Solving for the Initial Payment

A deferred contract pays three annual installments beginning two years from now. The first installment is $C_1$, and each subsequent installment is half the preceding one. A separate $200 payment is made with the final installment, at time 4. If the total present value is $1,080.32 at an effective annual interest rate of 25%, find $C_1$.

Method. Since the first regular payment is at time 2, $m=1$. The three payments are $C_1$ at time 2, $0.5C_1$ at time 3, and $0.25C_1$ at time 4. The terminal payment is $200 at time 4. With $v=0.8$, write the present-value equation, keeping the unknown regular-payment stream separate from the known terminal payment:

\begin{aligned} 1080.32 &=C_1(0.8)^2+0.5C_1(0.8)^3+0.25C_1(0.8)^4+200(0.8)^4\\ &=0.64C_1+0.256C_1+0.1024C_1+81.92\\ &=0.9984C_1+81.92. \end{aligned}

Subtract the terminal payment's present value and divide by the coefficient of $C_1$:

C_1=\frac{1080.32-81.92}{0.9984} =\frac{998.40}{0.9984} =1000.

The initial installment is $1,000. For a check, its three payments have present value $640+256+102.40=998.40$; adding the terminal payment's $81.92 gives the stated $1,080.32. The equation avoids the common error of subtracting the $200 terminal amount directly from the present value without discounting it.

Timing Traps and Practical Checks

The payment ratio and the interest rate affect the value in different ways. The ratio $r$ determines the size of successive installments; the discount factor $v$ accounts for when each installment is paid. The product $rv$ appears in the geometric sum because each successive installment is multiplied by $r$ and discounted one additional period by $v$.

A frequent error is to use the correct declining payments but assign them to the wrong dates. For a stream deferred by $m$ periods under this tutorial’s convention, the regular-payment dates are $m+1$ through $m+n$. The present-value factor on $C_1$ is therefore $v^{m+1}$, not $v^m$. For example, with a two-year deferral, the first payment is at time 3 and is discounted by $v^3$.

Another frequent error is to include a same-date terminal payment only once by treating it as though it replaced the last regular installment. The following incorrect calculation illustrates the mistake for the first worked example:

\text{Incorrect: }PV_0=800+320+128+122.88=1370.88.

This calculation drops the fourth regular installment, whose present value is $51.20. The correct calculation includes both that installment and the separate terminal amount:

\text{Correct: }PV_0=800+320+128+51.20+122.88=1422.08.

A final check is to compare values at a second focal date. If the interest rate is constant, the value at time $t$ is the value at time 0 accumulated by $(1+i)^t$, provided all cash flows are valued consistently. Alternatively, accumulate or discount each cash flow directly from its own date. The two approaches should agree.

Key takeaway: List the regular payment dates, calculate the declining installments, and value the terminal payment on its stated date as an additional cash flow. For a deferred stream, the first payment date determines the initial discount exponent; the word “terminal” does not determine whether the lump sum replaces or accompanies the final installment.

Check Your Understanding

For each problem, use the stated annual effective interest rate and treat a terminal payment as additional to the regular installments.

A three-payment stream begins at time 1. Its first payment is $800, each later payment is half the preceding payment, and a separate $100 is paid at time 3. At an effective annual rate of 25%, find the present value. (Answer: $1,049.60)
A deferred stream has three payments beginning at time 2. The first is $500, and each later payment is 80% of the preceding one. A separate $250 is paid at time 4. Find the present value at an effective annual rate of 25%. (Answer: $758.272, or $758.27 to the nearest cent)
At an effective annual rate of 25%, three payments of $400, $200, and $100 are made at times 1, 2, and 3, respectively. A separate $200 is also paid at time 3. Find the accumulated value at time 3. (Answer: $1,175)
A two-payment stream begins at time 3. Its first payment is $C_1$, and its second is $0.5C_1$. A separate $100 is paid at time 4. At an effective annual rate of 25%, the total present value is $614.40. Find $C_1$. (Answer: $800)
Three payments at times 2, 3, and 4 are $1,000, $500, and $250. An unknown terminal payment $X$ is also made at time 4. At an effective annual rate of 25%, the total present value is $1,200. Find $X$. (Answer: $492.1875, or $492.19 to the nearest cent)

What You'll Learn