Geometric Sequences in Context

Turn % increase/decrease into r; match 'start of year Y' or 'after n years' to the right term

In the exam for Edexcel International AS Maths · 5 real past-paper questions below

Percentages and the right term

You can already find a term and a sum of a GP. AS papers wrap the same GP in a story: a population, a price, a training plan. The hard part is turning the words into $a$ , $r$ and $n$ .

Rule

Increase by $p %$ each year: $r = 1 + \frac{p}{100}$ . Decrease by $p %$ : $r = 1 - \frac{p}{100}$ . So a 4% rise gives $r = 1.04$ , and a 15% fall gives $r = 0.85$ .

Formula

value after n years = a r^{n}

$a$ — the starting value, at time 0.
$r$ — the multiplier for one year.
$n$ — how many years have passed, so how many times you multiply.

Memory trick

Think of birthdays. On the day you are born you are 0, not 1. “After 5 years” is 5 birthdays, so $r^{5}$ . “The 5th year” counts year 1, 2, 3, 4, 5 from term 1, so only $r^{4}$ .

Worked example

1
A town uses 1200 megalitres of water in 2020. Its use rises by 3% each year. Find the use in 2027.
2
$r = 1.03$ — A 3% rise: multiply by 1.03 each year.
3
$2020 \to u_{1}, 2027 \to u_{8}$ — Count: 2020 is term 1, so 2027 is term 8.
4
$u_{8} = 1200 \times 1.0 3^{7} = 1475.8 \dots$ — The 8th term has $r$ seven times: about 1476 megalitres.

Common mistake

Using $r = 0.03$ for a 3% rise. That would shrink the value to 3% of itself. The multiplier is $1.03$ .

Key idea

Ask two questions: what is the multiplier, and how many times do I multiply?

Sum a run, then scale

Back to the town: 1200 megalitres in 2020, rising by 3% a year. Some questions add up only a run of years, such as “from 2023 to 2029 inclusive”, then turn the total into money. That run is a smaller GP: find its first term, count its terms, and sum it.

Worked example

1
The town pays 35 dollars per megalitre. Find the total cost of the water used from 2023 to 2029 inclusive.
2
$2023 \to u_{4} = 1200 \times 1.0 3^{3} = 1311.27 \dots$ — The run starts at term 4. This is the new $a$ .
3
$n = 10 - 4 + 1 = 7$ — 2029 is term 10. Count both ends.
4
$S = \frac{1311.27 \dots ( 1.0 3 ^{7} - 1 )}{1.03 - 1} = 10 047.58 \dots$ — Sum formula with the new $a$ and $n = 7$ .
5
$10 047.58 \dots \times 35 = 351665$ — Times the rate: about 351 665 dollars.

Common mistake

Counting $q - p$ terms. From 2023 to 2029 there are 7 years, not 6. You can check with $S_{10} - S_{3}$ : it gives the same 10 047.58.

Key idea

Keep full calculator values and round only the last line. The M mark is for the formula with your values in it, the A mark for the final number.

From AP to GP: least N

Some models change rule halfway. First the amounts drop by the same number each time (an AP). Later they drop by the same percentage (a GP). Treat the two parts separately, then add.

Worked example

1
A drill digs a well. On day 1 it digs 120 cm. Each day it digs 4 cm less than the day before, for the first 10 days. From day 11 the daily distances form a GP with ratio $r$ . On day 12 it digs 75 cm. Find $r$ , then the first day the well is deeper than 20 m.
2
$u_{10} = 120 + 9 (- 4) = 84$ — AP part: $a = 120$ , $d = - 4$ .
3
$S_{10} = \frac{10}{2} (120 + 84) = 1020$ — Depth after 10 days, in cm.
4
$84 r^{2} = 75$ — Day 11 is $84 r$ and day 12 is $84 r^{2}$ .
5
$r = \frac{75}{84} = 0.9449 \dots \approx 0.945$ — Take the positive root: distances are positive.
6
$1020 + \frac{84 r ( 1 - r ^{n} )}{1 - r} > 2000$ — GP part starts with $84 r$ on day 11. 20 m is 2000 cm.
7
$1 - r^{n} > 0.6801 \dots$ — Take 1020 from both sides, multiply by $1 - r$ and divide by $84 r$ .
8
$r^{n} < 0.3198 \dots$ — Rearrange.
9
$n l g r < l g 0.3198 \dots$ — Take lg of both sides, with the full value of $r$ . The power law brings $n$ down.
10
$n > 20.11 \dots$ — $l g r$ is negative because $r < 1$ , so dividing flips the sign.
11
$n = 21, N = 10 + 21 = 31$ — 21 GP days after the first 10: day 31.

Rule

Check the days either side: after day 30 the depth is about 1997 cm, after day 31 about 2022 cm. So day 31 is the first day deeper than 20 m.

Common mistake

Answering $n = 21$ . That counts only the GP days. The question asks for the day, so add the 10 AP days.

Common mistake

Starting the GP at 84. The 84 cm belongs to day 10, which is already in the AP total. Day 11 is $84 r$ .

Rule

“Show all steps in your working” or “you must show all stages”: write the inequality, the log line and the value of $n$ . A list of calculator totals with no method can score no marks.

Key idea

AP total, plus GP total, compared with the target. Logs bring $n$ down, then round up.

Fit N = ab^t, then compare

A GP over time can be written as $N = a b^{t}$ , where $t$ is the number of years. Here $a$ is the value at $t = 0$ and $b$ is the yearly multiplier. Given two values, you find $a$ and $b$ just like finding $a$ and $r$ from two terms.

Worked example

1
The number of fish in a lake is modelled by $N = a b^{t}$ . After 3 years there are 4600 fish. After 7 years there are 3300. Find $a$ and $b$ to 2 significant figures.
2
$a b^{3} = 4600, a b^{7} = 3300$ — Put each pair of values into the model.
3
$b^{4} = \frac{3300}{4600}$ — Divide: the $a$ cancels and $7 - 3 = 4$ .
4
$b = 0.9203 \dots \approx 0.92$ — Fourth root. The fish fall by about 8% a year.
5
$a = \frac{4600}{0.9203 \dots ^{3}} = 5901 \dots \approx 5900$ — Put $b$ back into $a b^{3} = 4600$ .
6
$N = 5900 \times 0.9 2^{t}$ — Write the model as an equation.

A second model says the number of frogs starts at 1500 and grows by 6% a year, so $M = 1500 \times 1.0 6^{t}$ . When are the two numbers equal?

Worked example

1
Find $T$ , to 1 decimal place, when the fish and frog numbers are equal.
2
$5900 \times 0.9 2^{T} = 1500 \times 1.0 6^{T}$ — Set the two models equal.
3
$(\frac{1.06}{0.92})^{T} = \frac{5900}{1500}$ — Collect the powers on one side and the numbers on the other.
4
$T l g (\frac{1.06}{0.92}) = l g (\frac{5900}{1500})$ — Take lg of both sides and use the power law.
5
$T = \frac{0.5947 \dots}{0.0615 \dots} = 9.66 \dots \approx 9.7$ — About 9.7 years.

Common mistake

Using $a b^{2}$ and $a b^{6}$ , as if year 3 were term 3. In $N = a b^{t}$ the power is $t$ itself, so 3 years means $b^{3}$ .

Common mistake

Writing $\frac{1.0 6 ^{T}}{0.9 2 ^{T}}$ as $(\frac{1.06}{0.92})$ with no power. Both have the same power $T$ , so the power stays on the bracket.

Key idea

Method marks go to making one power and taking lg; the A mark is for the value of $T$ . Write the lg line so the method is seen.

Key idea

In 1798 Thomas Malthus wrote that people multiply like a GP while food grows like an AP, so one day the GP must win. Comparing two growth models, as you just did, is the same idea.

Sum to infinity with trig

Sometimes $a$ and $r$ are written with $sin θ$ and $cos θ$ , and the sum to infinity is given. Use $sin^{2} θ = 1 - cos^{2} θ$ to get a quadratic in $cos θ$ .

Worked example

1
A GP has first term $5 sin^{2} θ$ and common ratio $3 cos θ$ . Its sum to infinity is 3. Show that $5 cos^{2} θ - 9 cos θ - 2 = 0$ , and find the exact second term.
2
$\frac{5 s i n ^{2} θ}{1 - 3 c o s θ} = 3$ — Use $S_{\infty} = \frac{a}{1 - r}$ .
3
$5 (1 - cos^{2} θ) = 3 - 9 cos θ$ — Replace $sin^{2} θ$ and multiply up.
4
$5 cos^{2} θ - 9 cos θ - 2 = 0$ — Collect on one side. “Show that” needs every line above.
5
$(5 cos θ + 1) (cos θ - 2) = 0$ — Factorise, like $(5 c + 1) (c - 2)$ .
6
$cos θ = - \frac{1}{5}$ — Reject $cos θ = 2$ : cos is never more than 1.
7
$r = 3 \times (- \frac{1}{5}) = - \frac{3}{5}, ∣ r ∣ < 1$ — Check the GP really has a sum to infinity.
8
$a = 5 (1 - \frac{1}{25}) = \frac{24}{5}$ — First term, using $sin^{2} θ = 1 - cos^{2} θ$ .
9
$a r = \frac{24}{5} \times (- \frac{3}{5}) = - \frac{72}{25}$ — The second term. Check: $\frac{24/5}{1 + 3/5} = 3$ .

Rule

For ratio $3 cos θ$ the sum exists only when $∣3 cos θ ∣ < 1$ , so $∣ cos θ ∣ < \frac{1}{3}$ . A root such as $cos θ = 0.5$ would give $r = 1.5$ : reject it.

Common mistake

Finding $θ$ on the calculator and using a rounded angle. The question wants an exact answer, so work with $cos θ = - \frac{1}{5}$ and $sin^{2} θ = \frac{24}{25}$ as fractions.

Key idea

Turn $sin^{2}$ into $1 - cos^{2}$ , solve the quadratic, then keep only the root with $∣ cos θ ∣ \leq 1$ and $∣ r ∣ < 1$ .

Which method? Read the question

The wording tells you which tool to use.

You see	You do
“increases by $p %$ each year”	$r = 1 + \frac{p}{100}$ ; a decrease gives $1 - \frac{p}{100}$
“after $n$ years”	$a r^{n}$
“in year $n$ ” or “start of 2027”, counting from term 1	$a r^{n - 1}$ ; list the first years to check
“from year $p$ to year $q$ inclusive”	New $a = u_{p}$ , $n = q - p + 1$ , then $S_{n}$
A cost or amount per unit	Multiply the total by the rate at the end
AP for a while, then GP	AP total + GP total; GP starts the term after the AP ends
“least”, “smallest” or “first” to pass a target	Inequality, lg both sides, flip if dividing by a negative, round up
“ $N = a b^{t}$ ” with two values	Divide to cancel $a$ , find $b$ , then $a$
“when the two numbers are equal”	Set the models equal, collect powers, take lg
$sin θ$ or $cos θ$ in $a$ or $r$ , with $S_{\infty}$	$sin^{2} θ = 1 - cos^{2} θ$ , quadratic, keep $∣ r ∣ < 1$

Common mistake

Giving the answer to the GP part only. Reread the question: it may ask for the day, the year or the total cost.

Key idea

Write $a$ , $r$ and what $n$ counts before any working. Most lost marks in these questions come from one wrong term, not hard maths.

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