Annuities & Series — Mathslive.ie

Section 1 of 9

Why we need series

In Compound Interest we had one lump sum sitting there growing. Now we have many payments — a save into the account every month, or a loan repayment every month.

The trick: each payment goes in at a different time, so each one grows (or gets discounted) for a different length of time. You add them all up. The sum turns out to be a geometric series — and we know the formula for that.

Must learn — the geometric sum:
$S_{n} = \dfrac{a(1 - r^{n})}{1 - r}$

$a$ = first term · $r$ = common ratio · $n$ = number of terms
(In the Tables.)

Every money question in this lesson eventually boils down to setting up a geometric series, spotting $a$, $r$ and $n$, then plugging in. The hard part is the set-up, not the algebra.

Section 2 of 9

Savings Annuity

Example: I will save €2,000 at the start of each year for 10 years. The AER is 3%. Find the value of my savings at the end of the 10th year.

(i) A rough check first

10 × €2000 = €20,000 (without interest)

So the answer must be a bit more than €20,000. Always do this check — it catches calculator errors.

(ii) Each payment grows for a different time

The €2000 paid at the start of year 1 sits for the full 10 years. The €2000 paid at the start of year 10 sits for just 1 year. Every payment in between sits for somewhere between those.

Year 1 payment: $2000(1.03)^{10}$

Year 2 payment: $2000(1.03)^{9}$

Year 3 payment: $2000(1.03)^{8}$

$\vdots$

Year 10 payment: $2000(1.03)^{1}$

$F = 2000(1.03)^{10} + 2000(1.03)^{9} + \cdots + 2000(1.03)^{1}$

(iii) Factor out and spot the geometric series

$F = 2000\!\left[\,1.03 + 1.03^{2} + 1.03^{3} + \cdots + 1.03^{10}\,\right]$

The bit in brackets is a geometric series. Read off the values:

$a = 1.03, \quad r = 1.03, \quad n = 10$

$S_{n} = \dfrac{a(1 - r^{n})}{1 - r} = \dfrac{1.03\!\left(1 - 1.03^{10}\right)}{1 - 1.03}$

$S_{n} = 11.808$

$F = 2000 \times 11.808$

F = €23,615.59

Bigger than €20,000, as predicted. The extra €3,616 is the interest earned across the 10 years.

TRY THIS

I save €1,500 at the start of each year for 8 years. The AER is 4%. Find the value of my savings at the end of the 8th year.

First payment grows for 8 years, last grows for 1 year. Sum is geometric with $a = 1.04$, $r = 1.04$, $n = 8$.

$F = 1500\!\left[1.04 + 1.04^{2} + \cdots + 1.04^{8}\right]$

$a = 1.04, \quad r = 1.04, \quad n = 8$

$S_{n} = \dfrac{1.04(1 - 1.04^{8})}{1 - 1.04} = 9.5828$

$F = 1500 \times 9.5828$

F = €14,374.19

F ≈ €14,374.19

TRY THIS

I save €600 at the start of each year for 12 years. The AER is 2.5%. Find the value of my savings at the end of year 12.

Same setup as the worked example. $a = 1.025$, $r = 1.025$, $n = 12$.

$F = 600\!\left[1.025 + 1.025^{2} + \cdots + 1.025^{12}\right]$

$a = 1.025, \quad r = 1.025, \quad n = 12$

$S_{n} = \dfrac{1.025(1 - 1.025^{12})}{1 - 1.025} = 14.140$

$F = 600 \times 14.140$

F = €8,484.27

F ≈ €8,484.27

Section 3 of 9

Start vs End

Read the question carefully. Where in the period each payment lands changes the time exponents.

Must learn:

Savings at start · first payment grows for $n$ full periods, last for 1 period.
Time exponents: $n,\ n-1,\ \ldots,\ 1$

Savings at end · first payment grows for $n-1$ periods, last for 0 periods.
Time exponents: $n-1,\ n-2,\ \ldots,\ 1,\ 0$

For loans the same rule applies — but the payments get discounted back (divided), not grown (multiplied).

(i) Worked example — Start of month

€350 is invested at the start of every month for 3 years. The AER is 2.6%. Find the value of the investment at the end of the 3rd year.

Number of payments: $3 \times 12 = 36$

Find the monthly rate: $(1+i)^{12} = 1.026$

$1+i = \sqrt[12]{1.026} = 1.002141$

First payment grows for 36 months, last for 1 month.

$F = 350\!\left[1.002141 + 1.002141^{2} + \cdots + 1.002141^{36}\right]$

$a = 1.002141, \quad r = 1.002141, \quad n = 36$

$S_{n} = \dfrac{1.002141(1 - 1.002141^{36})}{1 - 1.002141} = 37.462$

$F = 350 \times 37.462$

F = €13,111.83

Sanity check: 36 × 350 = €12,600 — answer is a bit more, looks right.

TRY THIS

€200 is invested at the end of every month for 2 years. The AER is 3.6%. Find the value at the end of year 2.

End of month → exponents run $23, 22, \ldots, 1, 0$. First convert 3.6% AER to monthly rate.

Payments: $2 \times 12 = 24$

Monthly rate: $1+i = \sqrt[12]{1.036} = 1.002952$

$F = 200\!\left[1 + 1.002952 + \cdots + 1.002952^{23}\right]$

$a = 1, \quad r = 1.002952, \quad n = 24$

$S_{n} = \dfrac{1 - 1.002952^{24}}{1 - 1.002952} = 24.833$

$F = 200 \times 24.833$

F = €4,966.51

F ≈ €4,966.51

Section 4 of 9

Saving for a Lump Sum

Same set-up, but now the question reverses. You know where you want to land (e.g. the price of a car) — what equal payment do you need each month?

The strategy:
Let the unknown payment be $P$. Set up the sum exactly as before — it'll factor as $P \times S_{n}$. Then solve $P \times S_{n} = \text{target}$ for $P$.

(i) Worked example

I save for a car in equal instalments over 2 years at 3% AER. The car will cost €12,000. What are my monthly savings at the start of each month?

Payments: $2 \times 12 = 24$

Monthly rate: $(1+i)^{12} = 1.03$

$1+i = \sqrt[12]{1.03} = 1.002$

Start of month → first payment grows for 24 months, last for 1 month.

$P(1.002)^{24} + P(1.002)^{23} + \cdots + P(1.002) = 12000$

$P\!\left[1.002 + 1.002^{2} + \cdots + 1.002^{24}\right] = 12000$

$a = 1.002, \quad r = 1.002, \quad n = 24$

$S_{n} = \dfrac{1.002466\!\left(1 - 1.002466^{24}\right)}{1 - 1.002466} = 24.754$

$24.754\, P = 12000$

$P = \dfrac{12000}{24.754}$

P = €484.77 per month

TRY THIS

I want to have €20,000 saved in 5 years for a deposit. The AER is 2.5%. What must I save at the start of each month?

$5 \times 12 = 60$ months. Convert 2.5% AER to monthly rate, then set up $P \cdot S_{60} = 20000$.

Monthly rate: $1+i = \sqrt[12]{1.025} = 1.002060$

$a = 1.002060, \quad r = 1.002060, \quad n = 60$

$S_{n} = \dfrac{1.002060(1 - 1.002060^{60})}{1 - 1.002060} = 63.927$

$P \times 63.927 = 20000$

P = €312.86 per month

P ≈ €312.86 per month

Section 5 of 9

Loan Repayments

With a loan the direction is reversed. The bank gives you the money now; you make a series of payments later. Each future payment is worth less today — so each one is discounted, not grown.

Key rule:
$\text{Loan now} = P = \dfrac{F}{(1+i)} + \dfrac{F}{(1+i)^{2}} + \cdots + \dfrac{F}{(1+i)^{t}}$

Each repayment $F$ gets pulled back to today's value.

(i) Worked example

€5,000 is borrowed for 3 years at 9% APR. It is paid back in equal instalments at the end of each month. What are the monthly payments?

Payments: $3 \times 12 = 36$

Monthly rate: $(1+i)^{12} = 1.09$

$1+i = \sqrt[12]{1.09} = 1.007207$

$5000 = \dfrac{F}{1.007207} + \dfrac{F}{1.007207^{2}} + \cdots + \dfrac{F}{1.007207^{36}}$

$5000 = F\!\left[\dfrac{1}{1.007207} + \dfrac{1}{1.007207^{2}} + \cdots + \dfrac{1}{1.007207^{36}}\right]$

The bracket is geometric. Read it off:

$a = \dfrac{1}{1.007207}, \quad r = \dfrac{1}{1.007207}, \quad n = 36$

$S_{n} = \dfrac{\dfrac{1}{1.007207}\!\left(1 - \left(\dfrac{1}{1.007207}\right)^{36}\right)}{1 - \dfrac{1}{1.007207}} = 31.609$

$5000 = 31.609\, F$

$F = \dfrac{5000}{31.609}$

F = €158.18 per month

Total paid back: 36 × 158.18 ≈ €5,695. The extra €695 is the interest the bank charged.

TRY THIS

I borrow a sum of money at 9% APR. I pay back €421.15 each month for 3 years at the end of each month. How much did I borrow?

Same setup but the unknown is $P$ now. Each €421.15 is discounted back; add them all up.

Monthly rate: $1+i = \sqrt[12]{1.09} = 1.007207$, $n = 36$

$P = 421.15\!\left[\dfrac{1}{1.007207} + \dfrac{1}{1.007207^{2}} + \cdots + \dfrac{1}{1.007207^{36}}\right]$

$a = \dfrac{1}{1.007207}, \quad r = \dfrac{1}{1.007207}, \quad n = 36$

$S_{n} = 31.609$

$P = 421.15 \times 31.609$

P = €13,312.14

P ≈ €13,312.14

TRY THIS

€10,000 is borrowed for 4 years at 6% APR, paid back monthly at the end of each month. Find the monthly payments.

$n = 48$ months. Find the monthly rate from APR first, then set up the discounted-payments series.

Monthly rate: $1+i = \sqrt[12]{1.06} = 1.004868$

$a = \dfrac{1}{1.004868}, \quad r = \dfrac{1}{1.004868}, \quad n = 48$

$S_{n} = 42.713$

$10000 = 42.713\, F$

F = €234.12 per month

F ≈ €234.12 per month

Section 6 of 9

The Amortisation Formula

For loans paid back in equal instalments — mortgages, car loans, anything — there is a clean formula. You will be asked to derive it on the LC, so let's see where it comes from.

(i) Derivation

€$P$ is borrowed over $t$ years. The APR is $i$. Equal annual repayments of $A$ are made. Find $A$ in terms of $P$, $i$ and $t$.

$P = \dfrac{A}{1+i} + \dfrac{A}{(1+i)^{2}} + \cdots + \dfrac{A}{(1+i)^{t}}$

$P = A\!\left[\dfrac{1}{1+i} + \dfrac{1}{(1+i)^{2}} + \cdots + \dfrac{1}{(1+i)^{t}}\right]$

$a = \dfrac{1}{1+i}, \quad r = \dfrac{1}{1+i}, \quad n = t$

$S_{n} = \dfrac{a(1 - r^{n})}{1 - r} = \dfrac{\dfrac{1}{1+i}\!\left(1 - \dfrac{1}{(1+i)^{t}}\right)}{1 - \dfrac{1}{1+i}}$

Tidy up. The denominator first:

$1 - \dfrac{1}{1+i} = \dfrac{(1+i) - 1}{1+i} = \dfrac{i}{1+i}$

$S_{n} = \dfrac{\dfrac{1}{1+i}\!\left(1 - \dfrac{1}{(1+i)^{t}}\right)}{\dfrac{i}{1+i}} = \dfrac{1}{i}\!\left(1 - \dfrac{1}{(1+i)^{t}}\right)$

$S_{n} = \dfrac{1}{i} \cdot \dfrac{(1+i)^{t} - 1}{(1+i)^{t}} = \dfrac{(1+i)^{t} - 1}{i(1+i)^{t}}$

$P = A \cdot \dfrac{(1+i)^{t} - 1}{i(1+i)^{t}}$

$A = \dfrac{P\, i\, (1+i)^{t}}{(1+i)^{t} - 1}$

Amortisation — mortgages and loans
(equal repayments at equal intervals)

$A = \dfrac{P\, i\, (1+i)^{t}}{(1+i)^{t} - 1}$

$A$ = repayment amount · $P$ = principal
(In the Tables.)

(ii) Worked example — Mortgage

I borrowed €250,000 over 30 years at 4% APR. Find my monthly repayments.

Monthly rate: $(1+i)^{12} = 1.04$

$1+i = \sqrt[12]{1.04} = 1.003274$

$i = 0.003274, \quad t = 30 \times 12 = 360$

$A = \dfrac{P\, i\, (1+i)^{t}}{(1+i)^{t} - 1}$

$A = \dfrac{250000 \times 0.003274 \times (1.003274)^{360}}{(1.003274)^{360} - 1}$

A = €1,183.25 per month

Total paid: 360 × 1183.25 ≈ €425,972. So nearly €176,000 of that is interest. Mortgages are expensive.

TRY THIS

I can afford €1,500 per month on a mortgage. The APR is 2.75%. How much can I borrow over 25 years?

Use the same formula but rearrange for $P$. First find the monthly rate from APR.

Monthly rate: $1+i = \sqrt[12]{1.0275} = 1.002263$, $i = 0.002263$

$t = 25 \times 12 = 300$

$1500 = \dfrac{P \times 0.002263 \times (1.002263)^{300}}{(1.002263)^{300} - 1}$

$1500 = P \times 0.004596$

P = €326,392.77

P ≈ €326,392.77

TRY THIS

I borrow €18,000 for a car over 5 years at 7.5% APR. Use the amortisation formula directly to find my monthly repayments.

Convert APR to monthly rate, then plug into $A = \dfrac{Pi(1+i)^{t}}{(1+i)^{t} - 1}$ with $t = 60$.

Monthly rate: $1+i = \sqrt[12]{1.075} = 1.006045$, $i = 0.006045$

$t = 60$, $P = 18000$

$A = \dfrac{18000 \times 0.006045 \times (1.006045)^{60}}{(1.006045)^{60} - 1}$

A = €358.58 per month

A ≈ €358.58 per month

Section 7 of 9

Pensions

Pensions are two problems glued together. Stage 1: you save into a pot for years. Stage 2: you receive payouts from the pot for more years.

The two stages:

Stage 1 — Accumulation. Money in monthly → grows → builds a pot.
Same set-up as a savings annuity (Section 2). Find $F$ = pot at retirement.

Stage 2 — Payout. The pot becomes the principal $P$. Monthly pension payments come out and the pot shrinks. Same set-up as a loan (Section 5).

The link: pot at end of Stage 1 = principal at start of Stage 2.

(i) Worked example

I am 30. I retire at 60 and will live to 80. I save €400 at the start of each month for the next 30 years. The AER is 3%. Find my monthly pension at the start of each month from 60 to 80.

Monthly rate: $(1+i)^{12} = 1.03$, $1+i = \sqrt[12]{1.03} = 1.002466$

Stage 1: build the pot (age 30 → 60)

Months: $30 \times 12 = 360$

First €400 grows for 360 months, last for 1 month.

$F = 400\!\left[1.002466 + 1.002466^{2} + \cdots + 1.002466^{360}\right]$

$a = 1.002466, \quad r = 1.002466, \quad n = 360$

$S_{n} = \dfrac{1.002466\!\left(1 - 1.002466^{360}\right)}{1 - 1.002466} = 580.140$

$F = 400 \times 580.140$

Pot at retirement: F = €232,056.12

Stage 2: take payouts (age 60 → 80)

Months: $20 \times 12 = 240$

Pot is the principal: $P = 232{,}056.12$

Let monthly pension be $F$. Start of month payouts:

$P = F + \dfrac{F}{1.002466} + \dfrac{F}{1.002466^{2}} + \cdots + \dfrac{F}{1.002466^{239}}$

$P = F\!\left[1 + \dfrac{1}{1.002466} + \cdots + \dfrac{1}{1.002466^{239}}\right]$

$a = 1, \quad r = \dfrac{1}{1.002466}, \quad n = 240$

$S_{n} = \dfrac{1 - (1/1.002466)^{240}}{1 - 1/1.002466} = 181.418$

$232{,}056.12 = 181.418\, F$

$F = \dfrac{232{,}056.12}{181.418}$

Monthly pension: F = €1,279.13

Sanity check: I paid in 400 × 360 = €144,000 and got back 1279.13 × 240 ≈ €307,000. Compound interest, working for you instead of against you.

TRY THIS

I am 25. I retire at 65 and will live to 85. I save €350 at the start of each month. The AER is 2.5%. Find my monthly pension at the start of each month from 65 to 85.

Stage 1: build the pot over 40 years (480 months). Stage 2: pot pays out for 20 years (240 months).

Monthly rate: $1+i = \sqrt[12]{1.025} = 1.002060$

Stage 1 (25 → 65, $n = 480$):

$S_{n} = \dfrac{1.002060(1 - 1.002060^{480})}{1 - 1.002060} = 819.742$

Pot: F = 350 × 819.742 = €286,909.78

Stage 2 (65 → 85, $n = 240$):

$S_{n} = \dfrac{1 - (1/1.002060)^{240}}{1 - 1/1.002060} = 189.594$

Pension: $F = \dfrac{286{,}909.78}{189.594}$

F = €1,513.29 per month

F ≈ €1,513.29 per month

Section 8 of 9

Net Present Value

Different question now. Should I invest in this project? A project costs money up front and then returns different amounts each year. Each year's return is worth less in today's money — we need to discount everything back.

The strategy:
Discount every cash flow back to today using $P = \dfrac{F}{(1+i)^{t}}$.
Add up all the present values (including the negative one — the initial cost).
That total is the Net Present Value (NPV).

Decision rule:
NPV $> 0$ → invest
NPV $\leq 0$ → do not invest

(i) Worked example

A new company is being set up. The cash flow is:

Year 0: $-50{,}000$ (start-up cost)
Year 1: $+5{,}000$
Year 2: $+15{,}000$
Year 3: $+35{,}000$

A discount rate of 6% is applied. Should the investor invest?

$i = 0.06$, $P = \dfrac{F}{(1+i)^{t}}$

Year 0: $-50{,}000$

Year 1: $\dfrac{5000}{1.06} = +4716.98$

Year 2: $\dfrac{15000}{1.06^{2}} = +13{,}349.95$

Year 3: $\dfrac{35000}{1.06^{3}} = +29{,}386.67$

NPV $= -50000 + 4716.98 + 13349.95 + 29386.67$

NPV = -€2,546.40

NPV is negative → do not invest.

Read the question: it's not about whether the project makes a profit on paper. €5,000 received in year 1 is not worth €5,000 today — it's worth less, because today's €5,000 could be earning interest. NPV asks: after discounting, do we come out ahead?

TRY THIS

I have a new app. It will cost €125,000 to develop. Returns over the next 3 years are €15,000, €55,000 and €70,000. The discount rate is 5%. Should I develop the app?

Discount each return back, sum with the negative start cost, check sign of NPV.

Year 0: $-125{,}000$

Year 1: $\dfrac{15000}{1.05} = +14{,}285.71$

Year 2: $\dfrac{55000}{1.05^{2}} = +49{,}886.62$

Year 3: $\dfrac{70000}{1.05^{3}} = +60{,}468.63$

NPV $= -125000 + 14285.71 + 49886.62 + 60468.63$

NPV = -€359.04 → do not develop

NPV ≈ -€359.04 → do not invest

TRY THIS

A project costs €80,000 to set up. Returns over the next 4 years are €20,000, €25,000, €30,000 and €25,000. The discount rate is 4%. Should the investor invest?

Discount each return back to today, sum with the negative start cost.

Year 0: $-80{,}000$

Year 1: $\dfrac{20000}{1.04} = +19{,}230.77$

Year 2: $\dfrac{25000}{1.04^{2}} = +23{,}114.05$

Year 3: $\dfrac{30000}{1.04^{3}} = +26{,}669.97$

Year 4: $\dfrac{25000}{1.04^{4}} = +21{,}369.88$

NPV = +€10,384.67

NPV $> 0$ → invest

NPV ≈ +€10,384.67 → invest

Section 9 of 9

Bonds

A bond is a loan you give a government or company. They pay you a fixed amount each year (the coupon), and at the end you get a final lump sum back. The question is: what's a fair price to pay for that bond today?

The strategy — two pieces:

A bond gives you both:
· a yearly instalment for $t$ years (annuity),
· a lump sum at the end (single payment).

Fair market value = present value of the instalments + present value of the lump sum.
Both are discounted back using $P = \dfrac{F}{(1+i)^{t}}$.

(i) Worked example

A bond gives a return of €100 per year for 5 years and a final return of €4,000 at the end of year 5. Find the fair market value at an APR of 3%.

Piece 1: the €100 instalments

$P_{1} = \dfrac{100}{1.03} + \dfrac{100}{1.03^{2}} + \cdots + \dfrac{100}{1.03^{5}}$

$P_{1} = 100\!\left[\dfrac{1}{1.03} + \dfrac{1}{1.03^{2}} + \cdots + \dfrac{1}{1.03^{5}}\right]$

$a = \dfrac{1}{1.03}, \quad r = \dfrac{1}{1.03}, \quad n = 5$

$S_{n} = \dfrac{\dfrac{1}{1.03}\!\left(1 - (1/1.03)^{5}\right)}{1 - 1/1.03} = 4.580$

P₁ = 100 × 4.580 = €457.97

Piece 2: the €4,000 lump sum

$P_{2} = \dfrac{4000}{1.03^{5}}$

P₂ = €3,450.44

Add them

Fair market value $= P_{1} + P_{2}$

$= 457.97 + 3450.44$

= €3,908.41

TRY THIS

A 10-year bond will return €8,000 at the end of the last year and €500 per year for the 10 years. The AER is 6%. Find the fair market value.

Two pieces: the €500 annual instalments (geometric series) and the €8,000 lump sum. Discount both, add.

Piece 1 — the €500 instalments:

$a = \dfrac{1}{1.06}, \quad r = \dfrac{1}{1.06}, \quad n = 10$

$S_{n} = \dfrac{(1/1.06)(1 - (1/1.06)^{10})}{1 - 1/1.06} = 7.360$

P₁ = 500 × 7.360 = €3,680.04

Piece 2 — the €8,000 lump sum:

$P_{2} = \dfrac{8000}{1.06^{10}}$ = €4,467.16

Total:

P₁ + P₂ = €8,147.20

Fair market value ≈ €8,147.20

End of Annuities & Series.

You now have the full Money HL toolkit. Every exam question is some combination of: compound interest, geometric series, and the present-vs-future direction.