IGCSE Add Maths 0606 / Topic 12 / IGCSE

Add the same step every time.

An arithmetic progression climbs or falls by the same amount at every step. Two formulas, one for the $n$th term and one for the sum of the first $n$ terms, answer almost every question about it: savings plans, rows of seats, training schedules and the sum of all the multiples of 7 between 100 and 300.

Full-screen labSequences and proportion lab → Topic 12All series lessons →

The key idea

An arithmetic progression (AP) has first term $a$ and adds the same common difference $d$ at every step, so $u_{n} = a + (n - 1)d$. The sum of the first $n$ terms is $S_{n} = \dfrac{n}{2}\big(2a + (n - 1)d\big) = \dfrac{n}{2}(a + l)$, where $l$ is the last term. A geometric progression (GP) MULTIPLIES by the same ratio instead. Three terms $x$, $y$, $z$ are consecutive terms of an AP exactly when the middle one is the average of the other two: $2y = x + z$.

Fig. 1: the sum of an AP by reversing it. The terms 2, 5, 8, 11, 14 (orange) and the same terms reversed (teal) fit together into 5 columns, each of height a + l = 16. So twice the sum is 5 × 16, and the sum is half of that, 40. In general 2S = n(a + l)

Section 01

Add the same difference.

A sequence is an arithmetic progression when the difference between each term and the one before it is always the same. That fixed amount is the common difference $d$: it is positive for a rising AP and negative for a falling one. In the lab, enter a few terms of a linear sequence and watch it find the difference and the $n$th term, then test whether a number is a term.

To reach the $n$th term you start at $a$ and take $n - 1$ steps of $d$:

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

Test	Example	Type
same DIFFERENCE $u_{n+1} - u_{n}$	3, 7, 11, 15 (add 4)	arithmetic (AP), $d = 4$
same RATIO $u_{n+1} \div u_{n}$	3, 6, 12, 24 (multiply by 2)	geometric (GP), $r = 2$, next lesson
neither is constant	1, 4, 9, 16	neither

Worked example AP, GP or neither?

Decide whether each sequence is arithmetic, geometric or neither. (a) 20, 17.5, 15, 12.5 (b) 2, 6, 18, 54 (c) 1, 3, 6, 10

(a) Differences: $-2.5, -2.5, -2.5$. The same every time, so an AP with $a = 20$ and $d = -2.5$.
(b) Differences: 4, 12, 36, not constant. Ratios: $6 \div 2 = 3$, $18 \div 6 = 3$, $54 \div 18 = 3$, so a GP with ratio 3.
(c) Differences: 2, 3, 4, not constant. Ratios: 3, 2, $\frac{5}{3}$, not constant. So neither.

Answer: (a) AP, $d = -2.5$ (b) GP, ratio 3 (c) neither

Worked example a and d from two terms

The 4th term of an AP is 17 and the 10th term is 41. Find the first term, the common difference and the 20th term.

$u_{4} = a + 3d = 17$ and $u_{10} = a + 9d = 41$.
Subtract: $6d = 24$, so $d = 4$.
Then $a = 17 - 3(4) = 5$.
$u_{20} = a + 19d = 5 + 76 = 81$.

Answer: $a = 5$, $d = 4$, $u_{20} = 81$

Your turn

How many terms are there in the AP 5, 9, 13, ..., 201?

Examiner trap

The $n$th term is $a + (n - 1)d$, not $a + nd$: the first term already counts as term 1, so the 20th term is only 19 steps from $a$. And find $d$ as a LATER term minus an EARLIER one, so a falling AP gets a negative $d$.

Section 02

Sum by pairing the ends.

Write the sum forwards and then backwards, one under the other (Fig. 1). Every column adds to the same total, the first term plus the last, and there are $n$ columns. So $2S_{n} = n(a + l)$. Putting $l = a + (n - 1)d$ gives the second form. Both are in the List of formulas.

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

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

Know the last term? Use $\frac{n}{2}(a + l)$: it is quicker.
Know $d$ but not $l$? Use $\frac{n}{2}\big(2a + (n - 1)d\big)$.
Number of terms from first to last: $n = \dfrac{l - a}{d} + 1$. Do not forget the $+1$.

Worked example Sum of 30 terms

Find the sum of the first 30 terms of the AP 4, 7, 10, ...

$a = 4$, $d = 3$, $n = 30$.
$S_{30} = \dfrac{30}{2}\big(2(4) + 29(3)\big) = 15(8 + 87)$.
$= 15 \times 95 = 1425$.

Answer: $1425$

Worked example Multiples of 7

Find the sum of all the multiples of 7 between 100 and 300.

First multiple above 100: $7 \times 15 = 105$. Last below 300: $7 \times 42 = 294$.
They form an AP with $a = 105$, $d = 7$, $l = 294$. Number of terms: $42 - 15 + 1 = 28$.
$S_{28} = \dfrac{28}{2}(105 + 294) = 14 \times 399 = 5586$.

Answer: $5586$

Your turn

The 5th term of an AP is 14 and the sum of the first 10 terms is 155. Find the first term and the common difference.

Examiner trap

In $\frac{n}{2}(a + l)$ the $l$ is the LAST TERM, not the number of terms. And count terms with the $+1$: from $7 \times 15$ to $7 \times 42$ there are 28 multiples, not 27.

Stage 1 · Learn

Check your understanding

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What is the common difference of the AP 13, 9, 5, 1, ...?

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Which sequence is an arithmetic progression?

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Find the $n$th term of the AP 5, 8, 11, ...

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Find the sum of the first 10 terms of the AP 1, 3, 5, ...

Section 03

Least n, contexts and three terms.

Exam questions often wrap an AP in a story: money saved each week, seats in each row, distance run each day. Identify $a$, $d$ and what is asked: ONE term (use $u_{n}$) or a RUNNING TOTAL (use $S_{n}$). When a total must exceed a value, $S_{n}$ gives a quadratic inequality in $n$, and $n$ must be a whole number.

Worked example A savings plan

Mia saves $20 in week 1 and each week saves $5 more than the week before. (a) How much does she save in week 12? (b) Find the least number of weeks for her total savings to exceed $1000.

(a) $a = 20$, $d = 5$: $u_{12} = 20 + 11(5) = 75$, so $75.
(b) $S_{n} = \dfrac{n}{2}\big(40 + 5(n - 1)\big) = \dfrac{n}{2}(5n + 35)$. Need $\dfrac{n(5n + 35)}{2} > 1000$.
Multiply by 2 and divide by 5: $n^{2} + 7n > 400$.
Try whole numbers: $n = 16$ gives $256 + 112 = 368$, too small; $n = 17$ gives $289 + 119 = 408 > 400$.
Check with the sums: $S_{16} = 8 \times 115 = 920$ and $S_{17} = \dfrac{17}{2} \times 120 = 1020$.

Answer: (a) $75 (b) 17 weeks

Worked example Three expressions in x

The expressions $x + 1$, $3x - 1$ and $4x + 2$ are consecutive terms of an AP. Find $x$ and the common difference.

In an AP the middle term is the average of its neighbours: $2(3x - 1) = (x + 1) + (4x + 2)$.
$6x - 2 = 5x + 3$, so $x = 5$.
The terms are 6, 14, 22: differences 8 and 8.

Answer: $x = 5$, $d = 8$

Your turn

Show that $(x - 1)^{2}$, $x^{2} + 1$ and $(x + 1)^{2}$ are consecutive terms of an AP for every value of $x$, and state the common difference.

Examiner trap

$n$ counts terms, so it must be a positive integer. If $n^{2} + 7n > 400$ holds from $n \approx 16.8$, the answer is 17, not 16: round UP, and check the whole numbers on either side. Also read whether the question is about the amount in ONE week ($u_{n}$) or the TOTAL so far ($S_{n}$).

Stage 2 · Exam

Exam-style questions

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The 3rd term of an AP is 12 and the 8th term is 32. Find the 20th term.

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A runner runs 3 km on day 1 and increases the distance by 0.5 km each day. Find the total distance run in the first 14 days.

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For the AP 2, 6, 10, ..., find the least value of $n$ for which $S_{n} > 500$.

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Find the sum of all the multiples of 6 between 50 and 200.

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Equal steps, summed by pairs.

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Go deeper · practice

Six original Cambridge-style questions

Telling an AP from a GP, a and d from two terms, the sum of multiples of 9, a term and a sum together with a least n, rows of theatre seats, and three expressions in k.

Test	Example	Type
same DIFFERENCE \(u_{n+1} - u_{n}\)	3, 7, 11, 15 (add 4)	arithmetic (AP), \(d = 4\)
same RATIO \(u_{n+1} \div u_{n}\)	3, 6, 12, 24 (multiply by 2)	geometric (GP), \(r = 2\), next lesson
neither is constant	1, 4, 9, 16	neither