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Decreasing annuities

Payments n, n−1, …, 1. The closed form (n − aₙ)/i, and the identity (Ia) + (Da) = (n+1)a that lets you check either one.

The formulas

Present value

$$(Da{)}_{\overline{n}|}=\frac{n-a_{\overline{n}|}}{i}$$

Accumulated value

$$(Ds{)}_{\overline{n}|}=\frac{n(1+i{)}^n-s_{\overline{n}|}}{i}$$

The checking identity

$$(Ia{)}_{\overline{n}|}+(Da{)}_{\overline{n}|}=(n+1)a_{\overline{n}|}$$

Where it comes from

1. A decreasing annuity is $n$ level annuities stacked: one of length $n$, one of length $n-1$, and so on.
2. ${\sum }_{k=1}^n{a}_{\overline{k}|}={\sum }_{k=1}^n\frac{1-v^k}{i}=\frac{n-a_{\overline{n}|}}{i}$.
3. Adding the increasing and decreasing versions pairs payment $k$ with payment $n+1-k$, so every period pays exactly $n+1$ — hence the identity.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• (Da)ₙ = (n − aₙ)/i: 'n minus the annuity', with a plain a this time — no dots.
• Always sanity-check with (Ia) + (Da) = (n+1)a. It costs ten seconds and catches a swapped formula.

Traps

• Using äₙ in the (Da) numerator — that factor belongs to (Ia).
• Reading 'decreasing by 1,000 a year for 10 years' as ending at 0 rather than at 1,000.

Related

• Increasing annuities
• Annuity-immediate

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