Sum of an Arithmetic Series

The sum of an arithmetic series: the sum formula, a sum in the middle, finding a and d, and finding how many terms (Higher).

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580 · 27 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What a Series Is (Higher)

A series is what you get when you add up the terms of a sequence. Arithmetic means the terms go up by the same amount.

Worked example

1
$4 + 9 + 14 + 19 + 24$ — five terms, going up by 5
2
$4 + 24 = 28, 9 + 19 = 28$ — pair the first with the last, the second with the second last
3
$\frac{5}{2} \times 28 = 70$ — 5 terms make two and a half pairs

Key idea

Every pair adds to the same total. That is the whole idea behind the formula.

The Sum Formula

Pairing gives a formula for the sum of the first n terms. Use it when there are too many terms to add one by one.

Formula

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

$a$ is the first term, $d$ is the common difference, $n$ is how many terms you add, and $S_{n}$ is their total.

Worked example

1
$4, 9, 14, \dots$ — The question: add the first 40 terms. Writing out 40 terms would take too long, so use the formula.
2
$a = 4, d = 5, n = 40$ — Read them off: the first term is a = 4, the gap 9 − 4 is d = 5, and you add 40 terms.
3
$S_{40} = \frac{40}{2} (2 \times 4 + (40 - 1) \times 5)$ — Put each number in its place. Write this line down: it earns a method mark.
4
$= 20 (8 + 39 \times 5)$ — Work out the easy parts: 40 ÷ 2 = 20, 2 × 4 = 8, and 40 − 1 = 39.
5
$= 20 (8 + 195) = 20 \times 203$ — Finish the bracket first: 39 × 5 = 195, then 8 + 195 = 203.
6
$= 4060$ — Multiply last: 20 × 203 = 4060.

Common mistake

Using n for the bracket. It is n − 1 lots of d: the 40th term is only 39 steps after the first.

Sum Using the Last Term

If you know the last term, there is a quicker formula. It is the pairing idea from the first card: every pair adds to first + last.

Formula

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

$a$ is the first term, $l$ is the last term, and $n$ is how many terms. There are $\frac{n}{2}$ pairs, each worth $a + l$ .

Add up all the multiples of 7 between 1 and 400

Worked example

1
$7, 14, 21, \dots, 399$ — List the start and the end. 400 ÷ 7 = 57.1…, so the last multiple below 400 is 57 × 7 = 399.
2
$a = 7, d = 7, l = 399$ — Each term is 7 more than the one before, so this is an arithmetic series.
3
$n = \frac{399 - 7}{7} + 1$ — Count the terms: the gap from first to last, divided by d, is the number of steps. Add 1 for the first term itself.
4
$n = 56 + 1 = 57$ — 392 ÷ 7 = 56 steps, so there are 57 terms.
5
$S_{57} = \frac{57}{2} (7 + 399)$ — Use the new formula: half of n, times first plus last.
6
$= \frac{57}{2} \times 406$ — Add inside the bracket: 7 + 399 = 406.
7
$= 57 \times 203 = 11571$ — Half of 406 is 203, which keeps the numbers whole. Then 57 × 203 = 11 571.

Common mistake

Forgetting the +1 when you count terms. From 7 to 399 is 56 steps of 7, but 57 numbers: count the first one too.

A Sum in the Middle

“Find the sum of the 21st to the 40th terms”: the first 40 terms are the ones you want plus the first 20. So take the first 20 away.

Worked example

1
$S_{40} = 4060$ — The sum of the first 40 terms of 4, 9, 14, …, worked out on the Sum Formula card.
2
$S_{20} = \frac{20}{2} (2 \times 4 + 19 \times 5)$ — Now the first 20 terms: the same a = 4 and d = 5, but n = 20, so the bracket has 20 − 1 = 19 lots of d.
3
$= 10 (8 + 95)$ — 20 ÷ 2 = 10, 2 × 4 = 8, and 19 × 5 = 95.
4
$= 10 \times 103 = 1030$ — Add inside the bracket, then multiply.
5
$S_{40} - S_{20} = 4060 - 1030 = 3030$ — Take the first 20 away. What is left is exactly terms 21 to 40.

Common mistake

Taking away the first 21. You want to keep the 21st term, so take away the first 20.

One single term works the same way. Some questions give a formula for the sum, not the terms. Find the 6th term when the sum of the first n terms is $n (2 n + 3)$ .

Worked example

1
$S_{n} = n (2 n + 3)$ — This adds up the first n terms. It is not the nth term, so you cannot just put n = 6 in.
2
$S_{6} = 6 (12 + 3) = 90$ — Put n = 6: the first 6 terms add up to 90.
3
$S_{5} = 5 (10 + 3) = 65$ — Put n = 5: the first 5 terms add up to 65.
4
$u_{6} = S_{6} - S_{5} = 90 - 65 = 25$ — The first 6 take away the first 5 leaves only the 6th term.

Find a and d

Some questions give a term and a sum. Write one equation for each, then solve them together.

Worked example

1
$u_{5} = 23 : a + 4 d = 23$ — The 5th term is a + (5 − 1)d: four steps of d after the first term. This is equation 1.
2
$S_{10} = 245 : \frac{10}{2} (2 a + 9 d) = 245$ — Put n = 10 into the sum formula. The bracket has 10 − 1 = 9 lots of d.
3
$5 (2 a + 9 d) = 245$ — 10 ÷ 2 = 5.
4
$2 a + 9 d = 49$ — Divide both sides by 5 to make the numbers smaller. This is equation 2.
5
$2 a + 8 d = 46$ — Double equation 1, so both equations start with 2a.
6
$(2 a + 9 d) - (2 a + 8 d) = 49 - 46$ — Take this away from equation 2. The 2a's cancel, leaving only d.
7
$d = 3$ — 9d − 8d = d, and 49 − 46 = 3.
8
$a + 4 \times 3 = 23, so a = 11$ — Put d = 3 back into equation 1: a + 12 = 23.
9
$5 (22 + 27) = 5 \times 49 = 245$ — Check in the sum: 2 × 11 = 22 and 9 × 3 = 27. It gives 245, so a = 11 and d = 3 are right.

Key idea

The nth term is $a + (n - 1) d$ . It turns every “the kth term is …” into an equation.

Key idea

Show clear working: write each equation on its own line. Each correct equation earns a method mark, even if a slip later spoils the answer.

Terms Given With a Letter

Some questions give the first three terms with a letter in them. In an arithmetic series the gaps are equal, so second − first = third − second. That gives an equation for the letter.

Worked example

1
$(5 x + 3), (2 x + 11), (x + 11)$ — The question: the first three terms. Find the sum of the first 20 terms.
2
$(2 x + 11) - (5 x + 3) = (x + 11) - (2 x + 11)$ — The gaps are equal. Keep the brackets so every part of the term is taken away.
3
$- 3 x + 8 = - x$ — Tidy each side: 2x − 5x = −3x and 11 − 3 = 8 on the left; x − 2x = −x and 11 − 11 = 0 on the right.
4
$8 = 2 x, so x = 4$ — Add 3x to both sides, then divide by 2.
5
$23, 19, 15$ — Put x = 4 into each term: 5 × 4 + 3, 2 × 4 + 11 and 4 + 11. The gaps match, so x = 4 is right.
6
$a = 23, d = - 4$ — The terms go down, so d is negative: 19 − 23 = −4.
7
$S_{20} = \frac{20}{2} (2 \times 23 + 19 \times (- 4))$ — Sum formula with n = 20. Put −4 in brackets so its minus is not lost.
8
$= 10 (46 - 76)$ — 2 × 23 = 46, and 19 × (−4) = −76, because plus times minus is minus.
9
$= 10 \times (- 30) = - 300$ — 46 − 76 = −30. A sum can be negative once the terms drop below zero.

Common mistake

Writing 2x + 11 − 5x + 3. Without brackets the +3 is not taken away. Bracket each term before you subtract.

One Sum Is k Times Another

Some questions link two sums, like “the sum of the first 12 terms is 3 times the sum of the first 6”. Write each sum with a and d, then turn the words into an equation.

Worked example

1
$S_{12} = 3 \times S_{6}, u_{4} = 13$ — The question. Find the first term a and the common difference d.
2
$S_{12} = \frac{12}{2} (2 a + 11 d) = 6 (2 a + 11 d)$ — Sum formula with n = 12. Leave a and d as letters, because you do not know them yet.
3
$S_{6} = \frac{6}{2} (2 a + 5 d) = 3 (2 a + 5 d)$ — Do the same with n = 6.
4
$6 (2 a + 11 d) = 3 \times 3 (2 a + 5 d)$ — The words say the first sum is 3 times the second. Write exactly that.
5
$12 a + 66 d = 18 a + 45 d$ — Multiply out both sides. On the right, 3 × 3 = 9, so it is 9(2a + 5d).
6
$21 d = 6 a$ — Take 12a and 45d from both sides, so a is on one side and d on the other.
7
$a = 3.5 d$ — Divide both sides by 6: 21 ÷ 6 = 3.5.
8
$a + 3 d = 13$ — The 4th term gives a second equation: three steps of d after a.
9
$3.5 d + 3 d = 13$ — Replace a with 3.5d, so only d is left.
10
$6.5 d = 13, so d = 2$ — Collect the d's, then divide by 6.5.
11
$a = 3.5 \times 2 = 7$ — Put d = 2 back into a = 3.5d.
12
$S_{12} = 6 (14 + 22) = 216, S_{6} = 3 (14 + 10) = 72$ — Check: 216 = 3 × 72, so the answer fits the question.

Key idea

If the question only asks for “a in terms of d”, stop at the line $a = 3.5 d$ . That line is the answer.

Find n from the Sum

If the sum is given and n is missing, the formula gives a quadratic in n.

Worked example

1
$5 + 8 + 11 + \dots = 440$ — The question: how many terms add up to 440? This time n is the unknown.
2
$a = 5, d = 3$ — The first term is 5, and the gap is 8 − 5 = 3.
3
$\frac{n}{2} (2 \times 5 + (n - 1) \times 3) = 440$ — Put a and d into the sum formula, keep n as a letter, and set it equal to the total.
4
$\frac{n}{2} (10 + 3 n - 3) = 440$ — 2 × 5 = 10, and 3 × (n − 1) = 3n − 3.
5
$\frac{n}{2} (3 n + 7) = 440$ — 10 − 3 = 7.
6
$n (3 n + 7) = 880$ — Multiply both sides by 2 to clear the half.
7
$3 n^{2} + 7 n - 880 = 0$ — Multiply out the bracket and take 880 from both sides, so the quadratic equals 0.
8
$(n - 16) (3 n + 55) = 0$ — Factorise. Check: 3n × (−16) = −48n and 55 × n = 55n, which add to 7n.
9
$n = 16 or n = - \frac{55}{3}$ — Set each bracket to 0. Only 16 works: n counts terms, so it must be a positive whole number.

Common mistake

Keeping the negative answer. You cannot add −18.3 terms.

Which Method? Read the Question

Write a, d and n before anything else. Then match the question to a method.

You see	You do
“Sum of the first n”	Put a, d, n into the sum formula
The last term, or “multiples of 7”	Count the terms, then $\frac{n}{2} (a + l)$
“Terms 21 to 40”	First 40 take away first 20
A formula for the sum, and “the 6th term”	$S_{6} - S_{5}$
A term and a sum	Two equations in a and d
Terms with a letter, like 2x + 1	Equal gaps: second − first = third − second
“One sum is 3 times another”	Write both sums with a and d, then make the equation
“How many terms”	Sum formula = total, solve for n
“The nth term”	Use $a + (n - 1) d$ , or see The nth Term of a Linear Sequence
A story: save $50, then $k more each year	Year 1 is a, the rise each year is d, the number of years is n

Key idea

Check your a and d by writing the first three terms. They should match the question.

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