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IB Maths AA SL Topic 1 — Number & Algebra Paper 1 & 2 🎯 Skill ~4 min practice

AA SL Arithmetic Series Sums Skills

An arithmetic series adds up terms with a constant difference. Two formulas in your booklet, one decision: pick the right one based on what the question gives you, and the rest is just plug-and-play.

The Method

AP: u1, u1 + d, u1 + 2d, u1 + 3d, … u1 = first term · d = common difference · n = number of terms
Formula 1

When you know u1 & d

Sn = n2(2u1 + (n − 1)d)
use when given first term and common difference
Formula 2

When you know u1 & un

Sn = n2(u1 + un)
use when given first term and last term

Three steps every time

  1. Identify what you have. Read the question and pick out u1, d, n, or un. d = any term minus the previous term.
  2. Choose the formula. Got u1 and d? Use Formula 1. Got u1 and un? Use Formula 2.
  3. Plug in carefully. Watch out for n vs (n − 1) — sloppy substitution is the #1 way to lose marks here.

Worked examples

WE 1 EASY

Find the sum of the first 20 terms of: 5, 8, 11, 14, …

step 1 — identify u₁ = 5, d = 8 − 5 = 3, n = 20 step 2 — Formula 1 (have u₁ and d) S₂₀ = 20/2 (2(5) + (20 − 1)(3)) = 10 (10 + 57) = 10 × 67 S₂₀ = 670 d = (any term) − (previous term) — pick any pair to find it!
WE 2 MEDIUM

An AP has first term 7 and 30th term 152. Find the sum of the first 30 terms.

step 1 — identify u₁ = 7, u₃₀ = 152, n = 30 step 2 — Formula 2 (have u₁ and uₙ) S₃₀ = 30/2 (7 + 152) = 15 × 159 S₃₀ = 2385 no need to find d — Formula 2 skips the (n−1)d step entirely!
WE 3 HARD

Find the sum of all multiples of 4 between 100 and 300 inclusive.

step 1 — identify the AP u₁ = 100, d = 4, last term = 300 step 2 — find n using uₙ = u₁ + (n−1)d 300 = 100 + (n − 1)(4) 200 = 4(n − 1) n − 1 = 50 → n = 51 step 3 — use Formula 2 (have u₁ and uₙ) S₅₁ = 51/2 (100 + 300) = 51/2 × 400 = 51 × 200 S₅₁ = 10 200 “between 100 and 300 inclusive” → both endpoints count. always check inclusive vs exclusive!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Pick the right formula before you start plugging in.
Q1 EASY Find the sum of the first 25 terms of: 3, 7, 11, 15, … Show answer ▼Hide answer ▲
u₁ = 3, d = 4, n = 25 → Formula 1 S₂₅ = 25/2 (2(3) + 24(4)) = 12.5 (6 + 96) = 12.5 × 102 S₂₅ = 1275
Q2 EASY An AP has u₁ = 12, u₅₀ = 502. Find S₅₀. Show answer ▼Hide answer ▲
u₁ and u₅₀ given → Formula 2 S₅₀ = 50/2 (12 + 502) = 25 × 514 S₅₀ = 12 850
Q3 MEDIUM An AP has u₁ = 20 and d = −3. Find the sum of the first 15 terms. Show answer ▼Hide answer ▲
Formula 1 (negative d works the same) S₁₅ = 15/2 (2(20) + 14(−3)) = 7.5 (40 − 42) = 7.5 × (−2) S₁₅ = −15
Q4 MEDIUM Find the sum of the first 100 positive even integers. Show answer ▼Hide answer ▲
2, 4, 6, …, 200 u₁ = 2, u₁₀₀ = 200, n = 100 → Formula 2 S₁₀₀ = 100/2 (2 + 200) = 50 × 202 S₁₀₀ = 10 100
Q5 HARD Find the sum of all multiples of 7 between 50 and 250. Show answer ▼Hide answer ▲
step 1 — find first & last multiples first multiple of 7 > 50: 56 last multiple of 7 < 250: 245 step 2 — find n 245 = 56 + (n − 1)(7) 189 = 7(n − 1) → n = 28 step 3 — Formula 2 S₂₈ = 28/2 (56 + 245) = 14 × 301 S₂₈ = 4214 “between” without “inclusive” — usually exclusive of endpoints!

⚠ Common mistakes

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Want the theory?

Read the full Arithmetic Sequences & Series notes for the link to the un formula, the proof of why these work, and applications to real-world problems like savings and loan repayments.

Need help with Arithmetic Series Sums?

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