IB Maths AI HLSequences & SeriesPaper 1 & 2common difference d~7 min read
Arithmetic Sequences & Series
An arithmetic sequence goes up (or down) by the same fixed amount each time — the common difference. Two formulae do almost everything: one for any term, one for the sum of the first n terms. Both are in the formula booklet.
📘 What you need to know
An arithmetic sequence has a constant difference between consecutive terms — the common difference d.
A negative d means the sequence decreases.
nth term: un = u1 + (n − 1)d.
Sum of the first n terms: Sn = n2(2u1 + (n − 1)d) or n2(u1 + un).
If u1 and d are unknown, two given terms give simultaneous equations.
If n is unknown, substituting into Sn often leads to a quadratic in n.
Arithmetic sequences
An arithmetic sequence is one where each term differs from the previous by a fixed amount, the common difference d. For 1, 4, 7, 10, … the difference is d = 3; a negative d gives a decreasing sequence.
The terms 4, 9, 14, 19, 24 rise by a constant d = 5. That fixed step is what makes the sequence arithmetic.
nth term of an arithmetic sequenceun = u1 + (n − 1)du1 = first term, d = common difference — given in the formula booklet
The sum of an arithmetic series
An arithmetic series is the sum of the terms of an arithmetic sequence. There are two formulae for Sn — use whichever fits the information given.
Sum of the first n termsSn = n2(2u1 + (n − 1)d)
Sn = n2(u1 + un)use the second when the last term un is known — both are in the booklet
The second form is quicker when you already know the last term; the first only needs u1 and d.
Working backwards
Many questions give you a term or a sum and ask for u1, d or n. Two given terms produce two equations — solve them simultaneously for u1 and d.
Finding n: if the number of terms is unknown, substitute into the Sn formula — this usually gives a quadratic in n. Solve it and keep only the positive whole-number answer.
🧠Recipe — arithmetic sequences and series
Identify u1 and d — d is the constant step between terms.
For a term, use un = u1 + (n − 1)d.
For a sum, use Sn = n2(2u1 + (n−1)d), or n2(u1 + un) if the last term is known.
If u1 or d is unknown, form two equations from the given terms and solve simultaneously.
If n is unknown, substitute into the formula and solve the resulting equation — often a quadratic.
Worked examples
WE 1
Finding a term
An arithmetic sequence has first term 5 and common difference 4. Find the 12th term.
use un = u1 + (n − 1)du12 = 5 + (12 − 1)(4)= 5 + 11 × 4 = 5 + 44u12 = 49it is (n − 1) lots of d — the first term already counts as term 1.
WE 2
Finding u₁ and d from two terms
The 3rd term of an arithmetic sequence is 17 and the 8th term is 42. Find the first term and the common difference.
write each term with the formulau3: u1 + 2d = 17u8: u1 + 7d = 42subtract the equations5d = 25 ⇒ d = 5u1 = 17 − 2(5) = 7u1 = 7, d = 5subtracting the two equations eliminates u1 and gives d at once.
WE 3
Summing an arithmetic series
Find the sum of the first 20 terms of the arithmetic sequence 3, 7, 11, 15, …
u1 = 3, d = 4, n = 20S20 = 202(2(3) + (20 − 1)(4))= 10(6 + 76) = 10 × 82S20 = 820use the first sum formula — it needs only u1, d and n.
WE 4
Finding n from a sum
An arithmetic sequence has first term 4 and common difference 3. How many terms are needed for the sum to reach 175?
set Sn = 175n2(2(4) + (n − 1)(3)) = 175n2(3n + 5) = 175 ⇒ 3n2 + 5n − 350 = 0solve the quadratic in nn = 10 or n = −35⁄3 (reject)n = 10 termsan unknown n in Sn gives a quadratic — keep only the positive whole number.
WE 5
Which term has a given value?
An arithmetic sequence is 2, 9, 16, 23, … Which term of the sequence is equal to 100?
u1 = 2, d = 7; set un = 1002 + (n − 1)(7) = 1007(n − 1) = 98 ⇒ n − 1 = 14100 is the 15th termsolving un = value gives n — it must be a positive integer.
WE 6
Full question: term, sum and a block of terms
An arithmetic sequence has first term 6 and common difference 5. (a) Find the 25th term. (b) Find the sum of the first 25 terms. (c) Find the sum of the 11th to the 25th terms inclusive.
(a) u25 = u1 + 24d= 6 + 24(5) = 126(b) S25 = 252(2(6) + 24(5))= 252(132) = 1650(c) 11th to 25th = S25 − S10S10 = 5(12 + 45) = 285; 1650 − 285(a) 126 · (b) 1650 · (c) 1365a block of terms is S of the last minus S just before the block starts.
💡 Top tips
The nth term uses (n − 1)d, not nd — the first term is already term 1.
Pick the sum formula to fit: use Sn = n2(u1 + un) when the last term is known.
Two given terms ⇒ two equations — subtract them to eliminate u1 and find d.
An unknown n inside Sn usually gives a quadratic — expect to solve one.
The sum of a block of terms is Slast − Sbefore.
âš Common mistakes
Using nd instead of (n − 1)d in the nth-term formula.
Finding the common difference the wrong way round — it is a later term minus an earlier one.
Forgetting the n2 factor in the sum formula.
Keeping a negative or non-integer n from a quadratic — only a positive whole number is valid.
Treating a multiplied or squared sequence as arithmetic — arithmetic means a constant added difference.
Next up: Geometric Sequences & Series — sequences where you multiply by a fixed common ratio instead of adding a fixed difference. The structure mirrors this note: an nth-term formula and a sum formula, plus a new idea, the sum to infinity.
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