Year 11 Extended · Unit 3 Sequences, series, formulae and proportion · Lesson 6 of 7

What happens when something is added every period and also changes?

Regular savings, loan repayments and repeated medicine doses all multiply what is there and then add or take away a fixed amount. This lesson shows each as a geometric series and uses the sum.

Number and AlgebraScientific calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 8: increase 800 by 5%.
    Answer840.
  2. From lesson 4: find 1 + 3 + 9 + 27.
    Answer40.
  3. From Year 9: work out 1.12.
    Answer1.21.
Words used in this lesson
deposit
money paid into an account. For example: paid at year end.
balance
the amount owed or held at a time. For example: after month 2.
repayment
money paid towards a loan. For example: equal each month.
dose
one amount of medicine taken at one time. For example: 400 mg.
ceiling
the level the amount approaches and does not pass. For example: the dashed line.

Will the medicine keep building up?

A patient in Kumasi takes a 500 mg tablet every 12 hours. By the time the next tablet is taken, the body has removed half of the drug that was there. The doctor wants to know whether the amount in the body will keep climbing, or settle.

Predict first: after many tablets, will the level just after a tablet keep rising for ever, or approach a value? Use the figure with a different fraction still in the body, and compare the level just after each dose with the dashed line.

40
1
100level just after the last dose (mg)
166.7ceiling (mg)

100 mg per dose, 40% still in the body at the next dose: just after dose 1 the level is 100 mg, and the ceiling is 166.7 mg.

Show me the steps
  1. 40% still in the body, 1 dose: the level just after the dose is 100 mg, and the ceiling is 166.7 mg.
  2. 40% still in the body, 3 doses: the level just after dose 3 is 100 × (1 + 0.4 + 0.16) = 156 mg, which is below the ceiling 166.7 mg.
  3. 80% still in the body, 5 doses: the level just after dose 5 is 100 × (1 + 0.8 + 0.64 + 0.512 + 0.4096) = 336.2 mg, and the ceiling is 500 mg, so the level is still well below it.
  4. Dose D, fraction f left at the next dose: the level after n doses is D(1 − fn) ÷ (1 − f), and the ceiling is D ÷ (1 − f).

Multiply by a factor, then add a fixed amount, and a geometric series appears. Unrolled, each model is a sum of powers of one factor:

Savings, P at the end of each year at rate i: Vn = P((1 + i)n − 1) ÷ i.
Loan L, interest i each period, payment P: Bn = L(1 + i)n − P((1 + i)n − 1) ÷ i.
Dose D with fraction f left at the next dose: D(1 − fn) ÷ (1 − f), with ceiling D ÷ (1 − f).

Why: the deposits are worth P + P(1 + i) + … + P(1 + i)n−1, a geometric sum with first term P and ratio 1 + i, so lesson 4 gives P((1 + i)n − 1) ÷ ((1 + i) − 1) = P((1 + i)n − 1) ÷ i. The dose sum D + Df + … + Dfn−1 is the same with ratio f.

The ceiling only exists because f is less than 1, as in the last lesson.

Two common slips. Using the fraction lost instead of the fraction left: if 30% leaves, 70% remains and f = 0.7, not 0.3. And counting deposits at the end of the year as if they had earned interest for the whole year: the last deposit has earned none, which is why the sum starts at (1 + i)0.

Worked examples

1. 1000 rupees is saved at the end of each year at 10% a year. What is the total after 3 years?

  1. The first deposit earns two years, the second one year and the third none: 1000 × (1.12 + 1.1 + 1).
  2. The formula gives 1000 × (1.13 − 1) ÷ 0.1 = 1000 × 0.331 ÷ 0.1 = 3310 rupees.

2. 10 000 pesos is borrowed at 2% interest a month and 1000 is repaid at the end of each month. Find the balance after 3 months.

  1. Month 1: 10 000 × 1.02 − 1000 = 9200. Month 2: 9200 × 1.02 − 1000 = 8384.
  2. Month 3: 8384 × 1.02 − 1000 = 7551.68 pesos. The formula agrees: 10 000 × 1.023 − 1000 × (1.023 − 1) ÷ 0.02 = 10 612.08 − 3060.40.

3. A patient takes 200 mg and 25% is still in the body at the next dose. Find the level just after the 3rd dose and the long-run ceiling.

  1. f = 0.25. After 3 doses: 200 × (1 + 0.25 + 0.0625) = 262.5 mg.
  2. The ceiling is 200 ÷ (1 − 0.25) = 266.67 mg, and the level approaches it.

Your go. The first step is given. A dose of 80 mg is taken and half is still in the body at the next dose. Find the level just after the 4th dose. The factor is f = 0.5.

  1. The level is 80 × (1 + 0.5 + 0.52 + 0.53).
  2. Work out the bracket and multiply by 80.
Check the answer

The bracket is 1 + 0.5 + 0.25 + 0.125 = 1.875, so the level is 80 × 1.875 = 150 mg.

With a calculator

Write the multiplier first: a rise of 7% a year is 1.07, and a drug with 35% still in the body keeps a factor of 0.35. Then use the power key, usually marked xy. To find (1.075 − 1) ÷ 0.07, work out 1.075 first, subtract 1, then divide.

Estimate first. Five years at 7% grows by a little more than 35%, so 1.075 should be a little over 1.4. If the display shows something close, the keys were right. Keep every decimal in the calculator until the end, and round only the final answer.

Check before you go on

1. A patient takes 150 mg. At the time of the next dose, 20% of the drug is still in the body. What is the level just after the 2nd dose?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A farmer in Kenya pays 4000 shillings into an account at the end of each year. The account earns 5% a year. What is the total after 3 years, to the nearest whole shilling?

3. A patient takes 400 mg of a medicine, and 30% of it is still in the body at the next dose. What is the level just after the 3rd dose, in mg?

Level 2 · Building

4. Each week 3 kg of a chemical is added to a pond, and 40% of the chemical present is lost by the next week. In the long run, how many kilograms are in the pond just after each addition?

5. A loan of 5000 naira has interest of 1% a month, added at the end of each month, and 500 is repaid straight after. What is the balance after 2 months, to one decimal place?

Level 3 · Stretch

6. A loan of 4200 is charged 10% interest a year. It is cleared exactly by two equal repayments, one at the end of each year. How much is each repayment?

7. A patient takes 50 mg and 60% is still in the body at the next dose. What is the smallest number of doses for which the level just after a dose is more than 110 mg?

Something to think about

A loan of 10 000 at 2% a month is repaid at 200 a month, which is exactly the interest in the first month. What happens to the balance, and what does a loan repaid at less than the interest do?

One way to think about it

The balance never falls: 200 is exactly 2% of 10 000, so each payment covers the interest and nothing more. With a payment below the interest the balance rises every month, because the interest added is more than the payment taken off.

Exit ticket

8. A saver pays 300 into an account at the start of each year, and the account earns 8% a year. Which calculation gives the total at the end of 4 years?