IB Maths AA HLTopic 1 ā Number & AlgebraPaper 1 & 2~9 min read
Arithmetic Sequences & Series
An arithmetic sequence is one where you add the same number to get from each term to the next. That fixed number is called the common differenced. Once you know d and the first term, you can jump straight to any term ā or sum any number of terms ā using formulas that are in the formula booklet. Most exam questions on this topic boil down to spotting which formula to use and what’s missing.
š What you need to know
Arithmetic = constant difference. Each term is the previous one plus d.
n-th term:un = u1 + (n ā 1)dā in the formula booklet.
Sum of first n terms ā two forms (both in the booklet): Sn = n2(2u1 + (n ā 1)d)
Sn = n2(u1 + un)
Use the first form when you have u1 and d; the second when you have u1 and the last term un.
If d > 0 the sequence increases. If d < 0 it decreases. If d = 0 every term is the same.
Harder questions often need simultaneous equations in u1 and d, or a quadratic in n when you’re asked for the number of terms.
What is an arithmetic sequence?
The defining property is simple: subtract any term from the next term and you always get the same value. That fixed value is the common differenced.
To check whether a sequence is arithmetic, compute (u2 ā u1), then (u3 ā u2), then (u4 ā u3). If they’re all equal, you have an arithmetic sequence ā and that common value is d.
The n-th term formula
To get from u1 to un, you take (n ā 1) steps of size d. That gives the formula:
n-th term of an arithmetic sequenceun = u1 + (n ā 1)dā in formula booklet
š¤ Why (n ā 1), not n?
Because u1 already counts as one term ā you’ve taken zero steps to get there. u2 requires one step, u3 requires two, and so on. The number of steps from u1 to un is one less than n.
Quick check: for the sequence 5, 9, 13, 17, 21 (u1 = 5, d = 4), the 10th term is u10 = 5 + (10 ā 1)(4) = 5 + 36 = 41.
Two formulas for the sum
Both sum formulas give the same answer ā they’re just rearrangements of each other. Choose whichever uses the information you’ve been given.
š
Form 1 ā using d
Sn = n2(2u1 + (n ā 1)d)
Use when you know u1, d, and n.
ā in formula booklet
šÆ
Form 2 ā using last term
Sn = n2(u1 + un)
Use when you know u1, the last term un, and n.
ā in formula booklet
š¤ Why two forms?
Form 2 is older ā Gauss famously used it as a child to add 1 + 2 + … + 100. Pair the first term with the last (1 + 100 = 101), the second with the second-last (2 + 99 = 101), and so on ā every pair sums to the same thing. Multiply by the number of pairs and divide by 2. Form 1 just substitutes un = u1 + (n ā 1)d into Form 2 ā same calculation in disguise.
In Paper 1 questions, Form 2 is often the cleaner choice ā particularly when the question phrases things in terms of “first term and last term”. In Paper 2, Form 1 is more flexible because it doesn’t require knowing the last term in advance.
Two harder patterns to recognise
Pattern A ā simultaneous equations in u1 and d
If a question gives you two pieces of information about specific terms (e.g. “the 4th term is 11 and the 9th term is 31”), you have two equations in u1 and d. Solve them simultaneously ā usually by subtracting one from the other to eliminate u1.
Pattern B ā quadratic in n
If a question gives you Sn as a number and asks you to find n, the sum formula becomes a quadratic in n. Expand, set equal to zero, and use the quadratic formula or factor. Reject negative solutions ā n is always a positive integer.
Recipe for Pattern B: use Sn = (n/2)(2u1 + (n ā 1)d), expand, multiply both sides by 2, rearrange to standard quadratic form (an2 + bn + c = 0), then solve.
Worked examples
WE 1
Find a specific term
An arithmetic sequence has first term 7 and common difference 4. Find the 15th term.
The third term of an arithmetic sequence is 11 and the eighth term is 26. Find the first term and the common difference.
Step 1: Set up the two equationsu3 = u1 + 2d = 11 … (1)u8 = u1 + 7d = 26 … (2)Step 2: Subtract (1) from (2)5d = 15 ā d = 3Step 3: Substitute backu1 + 2(3) = 11 ā u1 = 5u1 = 5, d = 3subtracting wipes out uā ā that’s the standard trick
WE 3
Find a sum using u1 and d (Form 1)
An arithmetic sequence has first term 8 and common difference 5. Find the sum of the first 20 terms.
Use Form 1 ā we have u1, d, and nSn = n2(2u1 + (n ā 1)d)S20 = 202(2(8) + 19(5))= 10 Ć (16 + 95) = 10 Ć 111S20 = 1110
WE 4
Find a sum using first and last terms (Form 2)
The first term of an arithmetic series is 4 and the 25th term is 100. Find the sum of the first 25 terms.
Use Form 2 ā we have u1, un, and nSn = n2(u1 + un)S25 = 252(4 + 100)= 252 Ć 104 = 25 Ć 52S25 = 1300Form 2 saves a step ā no need to find d when the last term is given
WE 5
Find n when given the sum (quadratic)
An arithmetic sequence has first term 5 and common difference 2. The sum of the first n terms is 320. Find n.
Step 1: Substitute into Form 1n2(2(5) + (n ā 1)(2)) = 320Step 2: Simplify the bracketn2(10 + 2n ā 2) = 320n2(2n + 8) = 320n(n + 4) = 320Step 3: Solve the quadraticn2 + 4n ā 320 = 0(n ā 16)(n + 20) = 0n = 16 or n = ā20Step 4: Reject the negative rootn = 16n must be a positive integer ā always reject negative solutions in counting problems
š” Top tips
Check it’s arithmetic first. Compute consecutive differences ā if they’re not all equal, the sequence isn’t arithmetic and these formulas don’t apply.
Pick the sum formula that matches what you have. Form 1 needs d; Form 2 needs the last term un. Don’t compute extra information you don’t need.
Two terms given ā simultaneous equations. The cleanest move is to subtract one from the other so u1 cancels.
Sum given, n unknown ā quadratic. Don’t just stare at it; multiply both sides by 2, expand, set to zero, solve.
Always check that any value of n you get is a positive integer. Decimal or negative answers mean check your working ā often a sign slip.
For Paper 1 (no calculator), keep things exact. Don’t decimalise mid-calculation ā leave fractions where useful.
Both n-th term and both sum formulas are in the formula booklet ā save brain space, keep the booklet open.
ā Common mistakes
Using n instead of (n ā 1) in the n-th term formula. The 10th term needs 9 steps from u1, not 10.
Mixing arithmetic with geometric. Arithmetic adds a fixed amount; geometric multiplies. Read the question carefully ā “common difference” vs “common ratio” tells you which.
Mis-setting up the sum formula. The 2 in 2u1 matters ā that’s a doubling, not “2 times something else”.
Forgetting to halve. Both sum formulas have an n/2 factor out front. Easy to drop under time pressure.
Keeping the negative root from a quadratic in n.n represents “number of terms” ā must be a positive integer.
Treating d = 0 as a special case. It’s still arithmetic ā every term equals u1, and the formulas still work.
Confusing the third term u3 with the 3rd-from-last. When questions give terms by position, count from the start unless told otherwise.
Arithmetic sequences are the most common type in IB papers ā at least one Paper 1 or Paper 2 question every year tests them. Master the simultaneous-equation technique and the quadratic-in-n setup, and you’ll handle any arithmetic question that turns up.
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