How do you choose between two roots?

Use the given condition, such as r > 1 or a positive ratio; never answer with both roots.

Why multiply the annuity by 1.03?

Each deposit is made at the start of the year and so earns one extra year of interest.

IB Maths AA: Sequences and Series (10 Challenge Items)

2026-11-08

Ten challenge items on sigma notation, recurring decimals as fractions, recovering a difference from a sum, choosing the valid root of a quadratic, symmetric arithmetic triples, an annuity due, and testing convergence.

Focus　Write the formula and known values first, and reject the invalid root of any quadratic.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	30
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: IB Maths AA: Sequences and Series
Level	IB Maths AA (HL)

02　Questions (10 items)

FoundationEvaluate the sum of (3k + 1) for k from 1 to 20.
FoundationFind 0.4 + 0.04 + 0.004 + … as a fraction.
FoundationAn arithmetic sequence has first term 4 and the sum of its first 20 terms is 650. Find the common difference.
FoundationDoes 1 + 1/3 + 1/9 + … converge? Find its sum if it does.
IntermediateA geometric sequence has first term 4 and the sum of its first 3 terms is 52, with r > 1. Find r.
IntermediateThree numbers form an arithmetic sequence with sum 24 and product 384. Find them.
Intermediate$3,000 is deposited at the start of each year at 3% per year. What is the total at the end of year 5 (2 d.p.)?
ChallengeAn arithmetic sequence has first term 5 and common difference 3. For which n does the sum of the first n terms first exceed 500?
ChallengeA geometric series has first term 6 and sum to infinity 9. Find the sum of the first 3 terms and compare it with 9.
Error redoA student applies S = a ÷ (1 − r) to a series with r = 3/2, gets a negative number and says the infinite sum is negative. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	The sum of 3k is 3 × 210 = 630, plus 20 ones, giving 650.	Split the constant from the k term; the sum of k from 1 to n is n(n + 1)/2.
2	With a = 0.4 and r = 0.1, S = 0.4 ÷ 0.9 = 4/9.	A recurring decimal can be written as an infinite geometric series; 0.444… is exactly 4/9.
3	650 = 10 × (8 + 19d), so 65 = 8 + 19d and d = 3.	Move n/2 to the other side first so the bracket is not expanded incorrectly.
4	r = 1/3 and \|r\| < 1, so it converges with S = 1 ÷ (1 − 1/3) = 3/2.	Convergence depends only on \|r\|, not on the size of the first term.
5	4(1 + r + r^2) = 52 gives r^2 + r − 12 = 0, so r = 3 or r = −4; since r > 1, r = 3.	The quadratic gives two roots; the condition r > 1 decides which one is valid.
6	The middle term is 8; then 8(64 − d^2) = 384 gives d = 4, so the numbers are 4, 8 and 12.	Writing the terms as a − d, a, a + d simplifies both the sum and the product.
7	3,000 × 1.03 × (1.03^5 − 1) ÷ 0.03 ≈ $16,405.23.	Each deposit at the start of a year earns one extra year of interest, so the whole sum is multiplied by 1.03.
8	S = n(3n + 7) ÷ 2; S17 = 493 and S18 = 549, so n = 18.	Check the values either side of the answer; rounding an inequality solution is unreliable.
9	r = 1/3; S3 = 6(1 − 1/27) ÷ (1 − 1/3) = 26/3 ≈ 8.67, which is 1/3 below 9, and that gap is the tail of the series.	With a positive first term and 0 < r < 1, every partial sum is below the sum to infinity.
10	When \|r\| > 1 the series diverges and has no sum, so the formula does not apply and the negative value means nothing.	The formula needs a geometric series, \|r\| < 1 and a finite first term; all three are required.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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