Should depreciation use 0.85 or 0.15?

Use 0.85: each year multiplies by 1 − 15%, so four years is a factor of 0.85^4.

How do you get a and d from a formula for S(n)?

S1 is the first term and S2 − S1 is the second term; their difference is d.

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

2026-11-08

Ten advanced items on compound interest and depreciation, negative differences and ratios, reversing the ratio, recovering a and d from S(n), and finding the first term above a target using powers of 2.

Focus　Name a and r or d before substituting; depreciation and interest are repeated multiplication.

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 (SL/HL)

02　Questions (10 items)

Foundation$5,000 is invested at 4% per year compound interest. What is its value after 3 years (2 d.p.)?
FoundationAn arithmetic sequence has first term 10 and common difference −3. Find the 15th term.
FoundationA geometric sequence has first term 3 and common ratio −2. Find the 5th term.
FoundationFind 1 + 3 + 5 + … + 19 (the sum of the first 10 odd numbers).
IntermediateA geometric sequence has first term 2 and common ratio 3. Find the sum of the first 6 terms.
IntermediateA geometric series has sum to infinity 24 and first term 6. Find the common ratio.
IntermediateIn the sequence 3, 6, 12, … which term first exceeds 1,000, and what is that term?
ChallengeThe sum of the first n terms of an arithmetic sequence is n^2 + 2n. Find the first term and the common difference.
ChallengeA new car worth $200,000 depreciates 15% per year. What is it worth after 4 years (to the nearest dollar)?
Error redoA student computes compound interest as 5,000 × (1 + 0.04 × 3) = 5,600. How far is this from the correct value?

03　Answers and one-line notes

#	Answer	One-line note
1	5,000 × 1.04^3 = 5,000 × 1.124864 = $5,624.32.	Compound interest multiplies by 1.04 each year; it does not add 4% of the original sum.
2	u15 = 10 + 14 × (−3) = 10 − 42 = −32.	The common difference may be negative, so keep its sign when substituting.
3	u5 = 3 × (−2)^4 = 3 × 16 = 48.	An even power of a negative ratio is positive, so the 5th term is positive.
4	This is arithmetic with a = 1, d = 2, n = 10, so S10 = 10/2 × (1 + 19) = 100.	The sum of the first n odd numbers is n^2, so 10^2 = 100 confirms the answer.
5	S6 = 2 × (3^6 − 1) ÷ (3 − 1) = 2 × 728 ÷ 2 = 728.	3^6 = 729 so the numerator is 728; the arithmetic must be done carefully.
6	6 ÷ (1 − r) = 24, so 1 − r = 0.25 and r = 0.75.	After solving, check \|r\| < 1; here 0.75 satisfies the condition.
7	3 × 2^(n − 1) > 1,000 needs 2^(n − 1) > 333.3; since 2^8 = 256 and 2^9 = 512, n − 1 = 9, so n = 10 and the term is 1,536.	Comparing successive powers of 2 is more reliable than trial and error on a calculator.
8	S1 = 3 so a = 3; S2 = 8 so the second term is 8 − 3 = 5 and d = 5 − 3 = 2.	Get a from S1, then the second term from S2 − S1; both steps must be shown.
9	200,000 × 0.85^4 = $104,401.25, about $104,401.	Depreciation multiplies by 0.85 four times; it is not 4 × 15% subtracted once.
10	The correct value is 5,624.32, so the gap is $24.32; 5,600 is simple interest, while compound interest reinvests each year’s interest.	The $24.32 gap is interest on interest, and it grows quickly with longer terms.

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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