Check the jump is always the same.
Subtract each term from the one after it.
a1
2
+3 →
a2
5
+3 →
a3
8
+3 →
a4
11
5 − 2 = 3 8 − 5 = 3 11 − 8 = 3
If the jumps were not all equal it would not be arithmetic, and this formula would not apply.
- Home
- Lesson Library
- Arithmetic sequences
Arithmetic Sequences
Same jump every time — one formula gets you any term you want.
Find the 10th term
2, 5, 8, 11, …
No need to write out all ten.
Here a = 2 and d = 3, and the nth term is a + (n − 1)d — so a₁₀ = 2 + 9(3) = 29. B uses 10d instead of 9d, the off-by-one that costs more marks than any other slip in sequences.
Write the formula and label the parts.
an = a1 + (n − 1)d
| Symbol | Means | Ours |
|---|---|---|
| an | the term you want | the 10th |
| a1 | the first term | 2 |
| n | which term | 10 |
| d | the common jump | 3 |
Everything you need is already in the question. The formula just tells you where to put it.
Why it is (n − 1) and not n.
Count the jumps, not the terms.
a1
2
1 →
a2
5
2 →
a3
8
3 →
a4
11
4 terms, but only 3 jumps
To reach term 4
3 jumps of 3
To reach term 10
9 jumps of 3
The off-by-one trap
Using 10 jumps instead of 9 gives 32, which is the 11th term. The first term is already there before any jump happens.
Drop the numbers in and solve.
a10 = 2 + (10 − 1)3
a10 = 2 + (9)(3)
10 − 1 = 9
a10 = 2 + 27
9 · 3 = 27
a10 = 29
add the first term back
Brackets first, then multiply, then add.
Adding them all up is a series.
A sequence is the list. A series is the total.
Sn = n · (a1 + an) ÷ 2
How many terms, times the average of the ends.
S10 = 10 · (2 + 29) ÷ 2
ten terms
S10 = 10 · 31 ÷ 2
2 + 29 = 31
S10 = 155
310 ÷ 2
The first ten terms add up to 155.
The sum is given instead. S₁₂ = 150 and a₁ = 3. Find a₁₂.
150 = 12 · (3 + a₁₂) ÷ 2
same formula, a₁₂ unknown
25 = 3 + a₁₂
150 ÷ 6
a₁₂ = 22
take away the first term
No last term? Put a₁ + (n − 1)d in for aₙ:
Sₙ = n · [2a₁ + (n − 1)d] ÷ 2
How many terms of 1 + 4 + 7 + … add up to 117?
117 = n · [2 + 3(n − 1)] ÷ 2
a₁ = 1, d = 3
234 = n(3n − 1)
times 2, tidy the bracket
3n² − n − 234 = 0
a quadratic in n
(n − 9)(3n + 26) = 0, so n = 9
n must be a positive whole number
Check: a₉ = 1 + 8(3) = 25 and 9 · (1 + 25) ÷ 2 = 117 ✓. The same form links a sum to d: if a₄ = 11 and S₈ = 100, then a₁ + 3d = 11 and 2a₁ + 7d = 25; double the first (2a₁ + 6d = 22) and subtract to get d = 3.
Backwards: how many terms are there?
Now the last term is given and n is the unknown. It goes into the same formula.
111 = 6 + (n − 1)(7)
d = 13 − 6 = 7
105 = (n − 1)(7)
take away the first term
n − 1 = 15
fifteen jumps
n = 16
one more term than jumps
Answering 15
15 is the number of jumps. The first term is already there before any jump, so the number of terms is always jumps + 1. Shortcut: n = (last − first) ÷ d + 1.
Three expressions in a row: make the jumps equal.
Consecutive terms have equal gaps: middle minus first equals last minus middle.
(2x + 1) − (x − 1) = (4x − 3) − (2x + 1)
first gap = second gap
x + 2 = 2x − 4
simplify each side
x = 6
solve
terms 5, 13, 21
both jumps are 8 ✓
The same idea, faster: the middle term is the average of its neighbours, so 2(middle) = first + last. Here 2(2x + 1) = 5x − 4 gives x = 6 again.
Answer what is asked
The question may want x, d or one of the terms. Substitute back: if the two jumps do not match, a sign slipped.
Variations: two given terms, fraction d, falling terms, a number that is not a term.
Two given terms. a₃ = 14 and a₉ = 32. Find the first term.
32 − 14 = 18
gap in values
9 − 3 = 6 jumps
gap in positions
d = 18 ÷ 6 = 3
values ÷ positions
a₁ = 14 − 2(3) = 8
walk back 2 jumps from a₃
What is different: neither a₁ nor d is given. Get d from the two terms first, then walk back to a₁. Check: a₉ = 8 + 8(3) = 32 ✓.
Fraction d. Find the 9th term of 12, 54, 2, …
d = 54 − 12 = 34
2 − 54 = 34 too ✓
a₉ = 12 + 8(34)
eight jumps
a₉ = 12 + 6 = 132
8 × 34 = 6
What is different: the method does not change, only the arithmetic. Multiply the jumps by d before adding.
Negative d. What is the first negative term of 50, 46, 42, …?
50 + (n − 1)(−4) < 0
d = 46 − 50 = −4
4(n − 1) > 50, so n − 1 > 12.5
move the jumps across
n − 1 = 13, n = 14
the next whole number of jumps
a₁₄ = 50 − 52 = −2
a₁₃ = 2 is still positive
What is different: the question is an inequality, not an equation. Round the number of jumps up, then check the terms on either side.
Not a term at all. Is 100 a term of 3, 7, 11, …?
100 = 3 + (n − 1)(4)
d = 4
n − 1 = 97 ÷ 4 = 24.25
not a whole number of jumps
No: a₂₅ = 99, a₂₆ = 103
100 falls between two terms
What is different: n must be a positive whole number. A fraction or a negative value of n means the number is not in the sequence.
Rounding n to a term
Rounding 24.25 and calling 100 “about the 26th term” is the classic slip. A term number is a count, so a fraction means no such term.
Given the sum formula: peel off one term.
Sₙ adds n terms and Sₙ₋₁ adds one fewer, so their difference is the nth term alone.
a₁ = S₁ = 1 + 4 = 5
a sum of one term is that term
S₂ = 4 + 8 = 12
the first two terms together
a₂ = S₂ − S₁ = 12 − 5 = 7
peel off a₁
d = 7 − 5 = 2
the sequence is 5, 7, 9, …
Check: S₃ = 9 + 12 = 21, and 5 + 7 + 9 = 21 ✓.
S₂ is not the second term
12 is the first two terms added together. Reading it as a₂ gives d = 7, and every later answer inherits the error.
On the exam
Sequence items rarely ask for a plain term. Expect two given terms (divide the gap in values by the gap in positions), “how many terms”, three expressions that must be consecutive, or a sum formula Sₙ to unpack. Ten-second check: a number is really a term only if (number − a₁) ÷ d comes out a whole number.
Check: count the sequence out to term 10.
1
2
2
5
3
8
4
11
5
14
6
17
7
20
8
23
9
26
10
29
the formula
2 + 9(3) = 29
matches the list
the sum
2 + 5 + … + 29 = 155
same as the series formula
Now you try
Practice problem
The problem
In an arithmetic sequence the 4th term is 14 and the 9th term is 34. What is the first term?
Subtract the two facts: from the 4th to the 9th term is five steps, so 5d = 34 − 14 = 20 and d = 4. Then a = 14 − 3(4) = 2. D is d itself, not the first term.
Hint: neither the first term nor d is given. Count the jumps between term 4 and term 9 to find d first.