Actuarial Mathematics for Modelling · Annuity and accumulation functions

Increasing and Decreasing Annuities: (Ia)n and (Da)n Formulas

Updated 11 October 2026 · Fact-checked

An increasing annuity pays amounts that rise by a fixed step, such as 1, 2, 3, … n. Its value is (Ia)n = (ä_n − n vⁿ) ÷ i. A decreasing annuity pays n, n−1, …, 1 and is worth (Da)n = (n − a_n) ÷ i. For geometric payments, use the ratio (1+g)÷(1+i) in a geometric series.

Understand Increasing and Decreasing Annuities

A level annuity pays the same amount each period. Many real cashflows do not. Payments may rise or fall by a fixed amount each year (an arithmetic progression), or grow by a fixed percentage (a geometric progression). The IAI tests both in CM1.

The standard symbol (Ia)n is the present value of an annuity-immediate paying 1 at the end of year 1, 2 at the end of year 2, and so on up to n at the end of year n. So (Ia)n = Σ t vᵗ for t = 1 to n. The symbol (Da)n pays n at time 1, n−1 at time 2, and so on down to 1 at time n. The letters I and D stand for increasing and decreasing.

You should be able to derive (Ia)n, not just memorise it. Multiply (Ia)n by (1+i). This shifts each term back one period: (1+i)(Ia)n = 1 + 2v + 3v² + … + n vⁿ⁻¹. Now subtract (Ia)n = v + 2v² + … + n vⁿ. What is left is i (Ia)n = 1 + v + … + vⁿ⁻¹ − n vⁿ = ä_n − n vⁿ. Divide by i to get the formula.

The decreasing annuity follows from a neat link. Adding the two patterns gives n+1 at every time from 1 to n. So (Ia)n + (Da)n = (n+1) a_n. Substitute (Ia)n and simplify to get (Da)n = (n − a_n) ÷ i.

For payments growing at a constant rate g, the present value is a geometric series with common ratio r = (1+g) ÷ (1+i). You do not need a special symbol. Sum the series, or use the formula below. Other patterns, such as payments that step up by P then Q, are handled by splitting them into a level annuity plus an increasing or decreasing annuity.

Key rules to remember

Increasing annuity-immediate: (Ia)n = (ä_n − n vⁿ) ÷ i; Payments 1, 2, …, n at times 1 to n. Here ä_n = (1 − vⁿ) ÷ d and v = 1 ÷ (1+i).
Increasing annuity-due: (Iä)n = (ä_n − n vⁿ) ÷ d; Payments 1, 2, …, n at times 0 to n−1. Same numerator as (Ia)n, but divide by d. Also (Iä)n = (1+i)(Ia)n.
Decreasing annuity-immediate: (Da)n = (n − a_n) ÷ i; Payments n, n−1, …, 1 at times 1 to n. Here a_n = (1 − vⁿ) ÷ i.
Decreasing annuity-due: (Dä)n = (n − a_n) ÷ d; Payments n, n−1, …, 1 at times 0 to n−1. Also (Dä)n = (1+i)(Da)n.
Link between increasing and decreasing: (Ia)n + (Da)n = (n+1) a_n; A quick check on your answers. Also useful to get one from the other.
Perpetuity increasing by 1 each year: (Ia)∞ = 1 ÷ (i d) = (1+i) ÷ i²; Payments 1, 2, 3, … at times 1, 2, 3, … forever. Needs i > 0.
Geometric annuity-immediate: PV = P × [1 − ((1+g) ÷ (1+i))ⁿ] ÷ (i − g), for i ≠ g; First payment P at time 1, then P(1+g), P(1+g)², … up to time n. If i = g, PV = n P ÷ (1+i).
Arithmetic step pattern: Payments P, P+Q, …, P+(n−1)Q at times 1 to n: PV = (P − Q) a_n + Q (Ia)n; Split into a level part and an increasing part. Q can be negative for a falling pattern.
Continuous increasing annuity: (Īā)n = (ā_n − n vⁿ) ÷ δ; Payment rate t at time t, so the total is ∫ t vᵗ dt from 0 to n. Here ā_n = (1 − vⁿ) ÷ δ.

How to solve Increasing and Decreasing Annuities questions

Use this method for any varying annuity question. The aim is to turn the cashflows into standard symbols, or into a geometric series, before you calculate.

1Write out the first few payments with their exact times. Note whether the first payment is at time 0 (due) or time 1 (immediate).
2Identify the pattern. Is the change a fixed amount (arithmetic) or a fixed percentage (geometric)? Does it increase or decrease?
3Match the pattern to (Ia)n or (Da)n. If the start value or step is not 1, rewrite the payments as a level part plus a multiple of (Ia) or (Da).
4If the payments are geometric, find r = (1+g) ÷ (1+i). Use the geometric formula, and check the special case i = g.
5Convert the interest rate if needed. Find i, v, d and δ from what the question gives, and keep full accuracy until the end.
6Calculate a_n or ä_n first, then the numerator, then divide by i or d. Use d for annuities-due.
7Adjust for timing if the first payment is not where the standard formula starts. Multiply by (1+i)ᵏ to move earlier, or by vᵏ to defer.
8Check with a quick sense test: compute the first one or two terms by hand, or use (Ia)n + (Da)n = (n+1) a_n. State your assumptions and the final answer with units.

Quickest way: Split, then use the standard formula

When to use it: Use this when payments rise or fall by a fixed amount and you are short of time. It also helps to catch the due versus immediate trap.

Write the first payment and the step: P and Q.
Find the standard symbol that matches, and put the amounts in terms of it: PV = (P − Q) a_n + Q (Ia)n. For a falling pattern with first payment P and fall Q, use PV = (P + Q) a_n − Q (Ia)n, which avoids (Da) if you prefer.
Compute v^n once, then a_n = (1 − vⁿ) ÷ i, then ä_n = a_n (1+i).
Use (Ia)n = (ä_n − n vⁿ) ÷ i or (Da)n = (n − a_n) ÷ i.
For a geometric pattern, skip symbols. Use PV = P [1 − rⁿ] ÷ (i − g) and check i ≠ g.
Do a one-line sanity check with the first payment. The PV should be less than the sum of undiscounted payments.

Common mistakes in Increasing and Decreasing Annuities

Using ä_n in (Da)n, or a_n in (Ia)n.
The two formulas look alike, and the due and immediate versions swap between numerator and denominator.
Fix: Remember: (Ia)n has ä_n − n vⁿ on top. (Da)n has n − a_n on top. Both divide by i for annuity-immediate. Test with n = 1: (Ia)1 = v and (Da)1 = v.
Dividing by i for an annuity-due.
Students copy the immediate formula and only change the first payment time.
Fix: Divide by d for (Iä)n and (Dä)n. Or compute the immediate value and multiply by (1+i).
Forgetting the n vⁿ term in (Ia)n.
Students remember ä_n ÷ i as the 'shape' and drop the correction for the final payment.
Fix: Rederive it in two lines by multiplying by (1+i) and subtracting. The n vⁿ term comes from the last payment of n.
Using a pattern starting at 5 or 10 as if it started at 1.
The question gives payments 5, 6, 7, … and students apply (Ia)n directly.
Fix: Split: payments P, P+1, … give (P − 1) a_n + (Ia)n. Check the first payment equals P.
Applying the geometric formula when i = g, or mixing up i − g with g − i.
The formula has a zero denominator in this case, and the sign is easy to flip when g > i.
Fix: If i = g, each term equals P ÷ (1+i), so PV = nP ÷ (1+i). If g > i, the formula still works: both top and bottom become negative.
Getting the number of terms or the time of the last payment wrong after a deferment.
Students shift the timing and forget to adjust v-powers.
Fix: List the times. For a pattern starting at time k+1, find the value at time k with the standard formula, then multiply by vᵏ.

Worked examples

Example 1

An annuity pays ₹1,000 at the end of year 1, ₹2,000 at the end of year 2, and so on, increasing by ₹1,000 each year, with the last payment of ₹10,000 at the end of year 10. Find the present value at 6% per annum effective.

Show the solution

The payments are 1,000 × (1, 2, …, 10), so PV = 1,000 (Ia)10 at i = 6%.
v¹⁰ = 1.06⁻¹⁰ = 0.558395.
a₁₀ = (1 − 0.558395) ÷ 0.06 = 7.36009.
ä₁₀ = a₁₀ × 1.06 = 7.80169.
n v^n = 10 × 0.558395 = 5.58395.
(Ia)₁₀ = (7.80169 − 5.58395) ÷ 0.06 = 2.21774 ÷ 0.06 = 36.9623.
PV = 1,000 × 36.9623 = ₹36,962 (to the nearest rupee).

Answer: ₹36,962 (approximately)

Example 2

A pension pays ₹10,000 at the end of year 1. Each later payment is 5% larger than the one before, for 3 payments in total. Find the present value at 10% per annum effective, using the geometric formula, and check by direct discounting.

Show the solution

Here P = 10,000, g = 5%, i = 10%, n = 3. Because i ≠ g, use the geometric formula.
The ratio r = 1.05 ÷ 1.10 = 0.954545. Then r³ = 1.157625 ÷ 1.331 = 0.869741.
PV = 10,000 × (1 − 0.869741) ÷ (0.10 − 0.05) = 10,000 × 0.130259 ÷ 0.05.
PV = 10,000 × 2.60519 = ₹26,052 (approximately).
Check: payments are ₹10,000, ₹10,500 and ₹11,025.
Discount: 10,000 ÷ 1.1 = 9,090.91. 10,500 ÷ 1.21 = 8,677.69. 11,025 ÷ 1.331 = 8,283.24.
Total = 9,090.91 + 8,677.69 + 8,283.24 = ₹26,051.84, which agrees.

Answer: ₹26,052 (approximately)

Exam tips

Show the standard notation first, such as 1,000 (Ia)10 at 6%, then the formula, then the numbers. Method marks depend on this.
Always check the timing in the question. 'Annuity-immediate' or 'in arrear' means the first payment is at time 1. Use d instead of i for due annuities.
In Paper B (computer-based), build the cashflows in a column and discount them directly. Compare the result with the closed-form formula as a check.
If the question gives a pattern that does not start at 1 or does not step by 1, split it into a level annuity plus an (Ia) or (Da) part before computing.
Look for questions that combine this topic with loans or equation of value, such as a loan repaid by increasing instalments. Set up the equation of value first, then solve for the unknown.

Practice questions from Annuity and accumulation functions

Increasing and Decreasing Annuities: frequently asked questions

What is the formula for (Ia)n and how is it derived?

For an annuity-immediate, (Ia)n = (ä_n − n vⁿ) ÷ i. Multiply Σ t vᵗ by (1+i), subtract the original sum, and you get i (Ia)n = ä_n − n vⁿ. This derivation is short enough to reproduce in the exam.

What is the difference between (Ia)n and (Iä)n?

(Ia)n pays 1, 2, …, n at times 1 to n. (Iä)n pays the same amounts one year earlier, at times 0 to n−1. So (Iä)n = (1+i)(Ia)n, and its formula divides by d instead of i.

How do I value an annuity whose payments grow geometrically?

Find r = (1+g) ÷ (1+i). Then PV = P (1 − rⁿ) ÷ (i − g) for payments P, P(1+g), … at times 1 to n. If i = g, every discounted payment equals P ÷ (1+i), so PV = nP ÷ (1+i).

Do I need to memorise (Da)n if I know (Ia)n?

You can get (Da)n from (Ia)n + (Da)n = (n+1) a_n. Even so, memorise (Da)n = (n − a_n) ÷ i, because it is faster and is often needed. Check it with n = 1, where it gives v.