Arithmetic Progressions in Problems

Turn facts about an AP into equations for a and d, find the least number of terms for a sum to pass a target, and solve AP problems in context.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 24 real past-paper questions below

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The AP toolkit

A theatre has 20 seats in the front row, and each row behind has 2 more seats. How many seats in row 15? How many seats in total? These are the two big AP questions: one term, and a running total.

In an arithmetic progression (AP) you add the same number each time. That number is the common difference $d$ . If you multiply by the same number instead, it is a geometric progression, which has its own lesson.

Formula

u_{n} = a + (n - 1) d

$a$ — the first term.
$d$ — the common difference: any term minus the term before it. It can be negative.
$n$ — the position of the term. The nth term has $n - 1$ jumps of $d$ after the first term.

Formula

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

$n$ — how many terms you add.
$a$ — the first term.
$d$ — the common difference.

If you know the last term, there is a shorter form. The sum is the number of terms times the average of the first and last.

Formula

S_{n} = \frac{n}{2} (a + l)

$a$ — the first term.
$l$ — the last term you add, which is $u_{n}$ .

Rule

0606 and 4037 print all three formulas in the List of formulas. 4PM1 prints only $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , so learn the nth term and the $\frac{n}{2} (a + l)$ form.

Worked example

1
A theatre has 20 seats in row 1, and each row has 2 more seats than the row in front. There are 15 rows. Find the number of seats in row 15 and the total number of seats.
2
$a = 20, d = 2, n = 15$ — Name the three values first.
3
$u_{15} = 20 + 14 \times 2 = 48$ — Row 15 is 14 jumps of 2 after row 1.
4
$S_{15} = \frac{15}{2} (20 + 48)$ — You know the first and last terms, so use the short form.
5
$= 7.5 \times 68 = 510$ — 510 seats in total.

Key idea

$u_{n}$ is one single term. $S_{n}$ is a running total. Read carefully which one the question gives you.

Two facts, two equations

Most AP questions give you two facts and want $a$ and $d$ . Each fact becomes one equation. Then you solve the pair together.

Steps

1
1. Write each fact using $u_{n}$ or $S_{n}$ , in terms of $a$ and $d$ .
2. Simplify each equation.
3. Solve the two equations together (substitute or eliminate).

Worked example

1
In an AP, the 7th term is twice the 3rd term. The sum of the first 12 terms is 270. Find $a$ and $d$ .
2
$a + 6 d = 2 (a + 2 d)$ — 7th term = 2 × 3rd term.
3
$a = 2 d$ — Expand to a + 6d = 2a + 4d, then collect: 2d = a.
4
$\frac{12}{2} (2 a + 11 d) = 270$ — Sum of the first 12 terms.
5
$2 a + 11 d = 45$ — Divide both sides by 6.
6
$4 d + 11 d = 45$ — Substitute a = 2d.
7
$d = 3, a = 6$ — 15d = 45, and a = 2 × 3.

Watch the wording with sums. "The next 10 terms" means the 11th to the 20th terms, which is $S_{20} - S_{10}$ .

Worked example

1
The sum of the first 10 terms of an AP is 155. The sum of the next 10 terms is 455. Find $a$ and $d$ .
2
$5 (2 a + 9 d) = 155 \Rightarrow 2 a + 9 d = 31$ — The first 10 terms.
3
$S_{20} = 155 + 455 = 610$ — The first 20 terms are the first 10 plus the next 10.
4
$10 (2 a + 19 d) = 610 \Rightarrow 2 a + 19 d = 61$ — Use the sum formula with n = 20.
5
$10 d = 30 \Rightarrow d = 3$ — Subtract the first equation from the second.
6
$2 a + 27 = 31 \Rightarrow a = 2$ — Put d = 3 back in.

Common mistake

Writing the 7th term as $a + 7 d$ . The first term already counts as one term, so the 7th term is $a + 6 d$ .

Common mistake

Using 455 as $S_{20}$ . "The next 10 terms" is only the second block. Add the first block to get $S_{20}$ .

Key idea

Write each equation clearly on its own line. Examiners give a mark for each correct equation, even if a later step slips.

How many terms?

Some questions ask how many terms there are, or the least number of terms for a sum to pass a target. Here the unknown is $n$ .

If you know the last term, use the nth term formula to find $n$ .

Worked example

1
Find the sum of the AP $7 + 11 + 15 + \dots + 203$ .
2
$7 + (n - 1) \times 4 = 203$ — The last term is the nth term.
3
$n - 1 = 49 \Rightarrow n = 50$ — 196 ÷ 4 = 49.
4
$S_{50} = \frac{50}{2} (7 + 203) = 5250$ — First and last terms known: short form.

For "least number of terms", write $S_{n}$ greater than the target. This gives a quadratic inequality in $n$ . Solve the matching equation, then round up.

Worked example

1
An AP has first term 5 and common difference 4. Find the least number of terms for the sum to be greater than 2000.
2
$\frac{n}{2} [10 + 4 (n - 1)] > 2000$ — Put a = 5 and d = 4 into the sum formula.
3
$n (2 n + 3) > 2000$ — Simplify the bracket: 10 + 4n − 4 = 4n + 6, and half of it is 2n + 3.
4
$2 n^{2} + 3 n - 2000 > 0$ — Expand and move 2000 across.
5
$n = \frac{- 3 + 16009}{4} = 30.88 \dots$ — Solve 2n² + 3n − 2000 = 0 with the quadratic formula. Ignore the negative root.
6
$n = 31$ — n is a whole number, so round up. Check: S₃₀ = 1890 and S₃₁ = 2015.

Common mistake

Rounding 30.88 down to 30. $S_{30} = 1890$ is still below 2000. For "greater than", always round up, then check.

Rule

On a calculator paper you may solve the quadratic on your calculator, but write the quadratic first. Forming it earns the method mark.

Key idea

Unknown $n$ ? Use $u_{n}$ if you know the last term. Use $S_{n}$ and an inequality if you know the target total.

Context and log terms

Problems in context are the same maths with words around them. Decide what the first term is, what the difference is, and what $n$ counts.

Worked example

1
A worker's salary goes up by the same amount every year. In year 3 it is 27 600 dollars and in year 8 it is 33 600 dollars. Find the total salary for the first 10 years.
2
$a + 2 d = 27600$ — Year 3 is the 3rd term.
3
$a + 7 d = 33600$ — Year 8 is the 8th term.
4
$5 d = 6000 \Rightarrow d = 1200$ — Subtract the first equation from the second.
5
$a = 27600 - 2400 = 25200$ — Put d back into the first equation.
6
$S_{10} = 5 (50400 + 10800) = 306000$ — Use the sum formula with n = 10.

Some APs are made of logs. Recall that $ln 9 = ln 3^{2} = 2 ln 3$ . So each term below is the one before plus $ln 3$ .

Worked example

1
Show that $ln 3 + ln 9 + ln 27 + \dots$ is an AP, and find the sum of the first 20 terms in the form $k ln 3$ .
2
$ln 3, 2 ln 3, 3 ln 3, \dots$ — Power law on each term: 9 = 3², 27 = 3³.
3
$d = ln 3$ — Each term is the one before plus ln 3, so it is an AP.
4
$S_{20} = \frac{20}{2} [2 ln 3 + 19 ln 3]$ — a = ln 3 and d = ln 3.
5
$= 10 \times 21 ln 3 = 210 ln 3$ — Collect the ln 3 terms.

Rule

On 0606 Paper 1 there is no calculator. Leave log answers exact, like $210 ln 3$ .

Common mistake

Counting from 0 in a context. Year 1 is the first term, so year 8 is $a + 7 d$ , not $a + 8 d$ .

Key idea

Strip away the story: find $a$ , $d$ and what $n$ counts. Then it is the same AP you already know.

Which method? Read the question

The wording tells you which formula and which method to use.

You see	You do
A term given, like "the 7th term is…"	$u_{n} = a + (n - 1) d$
"The sum of the first n terms is…"	$S_{n} = \frac{n}{2}$ $[2 a + (n - 1) d]$
First and last terms known	$S_{n} = \frac{n}{2} (a + l)$
Two facts, find $a$ and $d$	Two equations, solve together
"The next 10 terms"	$S_{20} - S_{10}$
"Least number of terms"	Inequality in $n$ , solve, round up
Terms made of logs	Power law, then find $d$

Common mistake

Mixing up $u_{n}$ and $S_{n}$ . One is a single term, the other is a total.

Key idea

Name $a$ , $d$ and $n$ first. Every AP question gets easier once those three are clear.

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