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Modelling with series

Use arithmetic and geometric series for totals, budgets, capped growth, bouncing distances and regular payments. Learn payment timing with a manual model and worked practice.

Before you startArithmetic and geometric sums, sum to infinity and logarithms

01 / What quantity must be added?

A term and a cumulative total answer different questions.

If uₙ is the distance travelled on day n, then Sₙ = u₁ + u₂ + … + uₙ is the distance travelled over the first n days. The question “how far on day 8?” asks for u₈; “how far by the end of day 8?” asks for S₈.

Write the first few amounts actually being added. Their units and dates often reveal a wrong first term, a missing journey or an extra interest period before any algebra is needed.

02 / Totals with a fixed increase

Add all stages, not just the final one.

A display has 15 rows, with 18 lights in the first row and 3 more in each later rowWorked example

u₁₅ = 18 + 14(3) = 60

This is the number in the last row.

S₁₅ = 15(18 + 60)/2 = 585

The average of first and last, multiplied by the number of rows, gives the total.

A monthly cost starts at £30 and rises by £5 each month. How many whole months fit a £500 budget?Worked example

Sₙ = n[60 + 5(n − 1)]/2 = 5n(n + 11)/2

The budget applies to the cumulative cost.

S₉ = £450; S₁₀ = £525

All costs are positive, so these adjacent checks identify the last affordable month.

Nine whole months are affordable

There is £50 left, less than the £75 tenth-month cost.

If the terms can be negative, a cumulative sum need not increase; adjacent checks alone then need a separate monotonicity argument.

03 / Add a capped sequence in parts

Split where the rule changes.

Annual output is min(240 + 35(n − 1), 500). Find the first 12 years’ total.Worked example

Years 1–8 are 240, 275,…, 485

Year 9 is the first capped year: its uncapped prediction is 520.

S₈ = 8(240 + 485)/2 = 2900

Use an arithmetic sum for the uncapped years.

Years 9–12 contribute 4 × 500 = 2000

Count the capped years inclusively.

Total = 2900 + 2000 = 4900 items

The cap is included once for each of the last four years.

Using one arithmetic formula for all 12 years would count output the cap forbids. Writing the two index ranges prevents an omitted or double-counted boundary year.

04 / Totals with a fixed percentage increase

Find the first amount in the sum and its ratio.

A route plan starts at 20 km on day 1 and rises by 10% each dayWorked example

uₙ = 20(1.1)^(n − 1)

This describes the planned distance on one day.

Sₙ = 20[(1.1)ⁿ − 1]/0.1

This describes the planned total after n complete days.

S₇ = 189.74342 km; S₈ = 228.717762 km

Keep full precision until the final result.

A percentage increase in daily distance does not mean the cumulative total itself grows by that same percentage each day. Derive the sum from the individual terms.

05 / Stop when the target is reached

The final day may use only part of the planned term.

Use the same 20 km, 10%-increase plan, but stop after 220 kmWorked example

S₇ < 220 ≤ S₈

The target is reached during day 8.

Final-day distance = 220 − S₇ = 30.25658 km

Only the remaining distance is needed, not the full planned day-8 amount.

Day 7 distance = 20(1.1)⁶ = 35.43122 km

The earlier full-day distances increase.

The greatest actual daily distance is on day 7

Its 35.43122 km exceeds the shortened final day’s 30.25658 km.

“Eight days are needed” means seven full planned days plus part of the eighth here. A fraction of a day requires an extra speed or within-day pacing assumption; the daily sequence alone does not determine it.

06 / Bounce heights and total distance

Every completed rebound includes an upward and a downward trip.

A ball is dropped from 6 m. Each rebound reaches 60% of the preceding height.Worked example

Rebound n reaches hₙ = 6(0.6)ⁿ

The initial drop is not a rebound. The fourth rebound reaches 0.7776 m.

The sixth ground impact follows five complete rebounds

The initial drop ends at impact 1.

Distance = 6 + 2[6(0.6) + … + 6(0.6)⁵]

Count the initial drop once and each completed rebound twice.

= 6 + 12(0.6)[1 − (0.6)⁵]/(1 − 0.6) = 22.60032 m

The last rebound counted is number 5.

To ground impact k: D = H + 2Hr(1 − r^(k − 1))/(1 − r)
Infinite idealised travel: D∞ = H + 2Hr/(1 − r), for 0 < r < 1

Here the idealised infinite distance is 24 m. If the endpoint is the top of a rebound, its final upward journey is counted only once. Real balls eventually stop bouncing, so the infinite result belongs to the idealised model.

07 / Payments at the end of each year

The latest payment has earned no interest at the valuation time.

Deposit £250 at each year-end for six years; annual interest is 4%. Value immediately after the sixth deposit.Worked example

The first deposit earns five years’ interest; the last earns none

List the contribution of each payment at the same final date.

Balance = 250(1.04)⁵ + 250(1.04)⁴ + … + 250

In increasing powers, this is a geometric series with first term 250.

= 250[(1.04)⁶ − 1]/0.04 = £1658.24

Round the final sum, not each intermediate power.

End-year payments P, multiplier r, n years:
Bₙ = P(rⁿ − 1)/(r − 1), for r ≠ 1

This model assumes a constant rate, no fees or withdrawals and no intermediate rounding. When r = 1, there is no interest and the balance is simply nP.

Give each payment its own growth timeExplore

Pay £100 every year. Value the balance at the end of the final year, after that year’s interest and any end-year payment.

Year 1: 3 interest periods → £115.76
Year 2: 2 interest periods → £110.25
Year 3: 1 interest period → £105.00
Year 4: 0 interest periods → £100.00

Four end-year payments total £431.01 at the end of year 4.

Paid in: £400.00. Interest: £31.01.

Bars compare the final contribution of each payment on a shared scale. Calculations retain full precision until the final total is rounded. Individual rows are displayed to the nearest penny, so their rounded sum may differ by a penny.

Payment time Annual increase Number of years 4

Watch each payment reach the valuation date

Pause, replay or seek freely. The notes explain the same idea and stay in view.

08 / Payments at the start of each year

At the same final date, every payment gets one extra interest period.

Instead deposit £250 at each year-start for six years, and value at the end of year 6.Worked example

The first payment earns six years’ interest; the last earns one

The payment dates have moved one year earlier.

Balance = 250(1.04)⁶ + … + 250(1.04)

There is no term with exponent zero.

= 1.04 × 250[(1.04)⁶ − 1]/0.04 = £1724.57

Multiply the unrounded end-payment expression by 1.04.

Start-year payments, valued at the final year-end:
Bₙ = Pr(rⁿ − 1)/(r − 1), for r ≠ 1

If valuation is immediately after the last start-year payment instead, the exponents are n − 1 through 0. The date of valuation matters as much as the payment label.

09 / When will regular payments reach a target?

Solve for the number of complete payment-and-interest cycles.

Pay £200 at each year-start, earn 5% annually, and assess the balance each year-end. First reach £2900.Worked example

Bₙ = 200(1.05)[(1.05)ⁿ − 1]/0.05

The latest payment earns one year of interest.

(1.05)ⁿ ≥ 1 + 2900(0.05)/[200(1.05)]

Rearrange before taking logarithms.

n ≥ 10.7606, so test n = 11

The first candidate whole year is obtained by rounding upward.

Pay P at each year-start for n years at multiplier r. Write the balance immediately after the last payment, before the final year’s interest.

Hint

The most recent payment has exponent zero at this date.

Worked solution

Balance = P[r^(n − 1) + … + r + 1] = P(rⁿ − 1)/(r − 1) for r ≠ 1, or nP when r = 1. The different valuation date removes the extra factor r.

11 / Recap

Every term must represent the right amount at the right time.

Distinguish an individual amount from a cumulative sum.
Split capped models into the correct index ranges.
Check whether the final stage is shortened.
Count rebound journeys and the initial drop separately.
Value every payment at the same final date.
Check adjacent integer periods for a target.
State rounding and modelling assumptions.

Review the chapter’s other lessons →

Section 1 of 11 · What quantity must be added?