This IB math sequences series practice question will test your understanding of geometric series — one of the most important topics in the AA SL and HL course. Sequences and series appear every year on IB exams, and getting comfortable with common ratio, partial sums, and convergence is essential for earning full marks. Give this question a genuine attempt before scrolling down to the solution. You’ll learn far more by struggling with it first!
⏱ 8–10 minutes
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This Week’s Challenge — IB Math Sequences Series Practice
Paper 2 style (calculator allowed)
📝 Question of the Week
A geometric sequence has first term \( u_1 = 48 \) and common ratio \( r = \frac{3}{4} \).
(a) Find the value of \( u_5 \), the fifth term of the sequence.
[2 marks]
(b) Find the sum of the first 10 terms of the sequence, \( S_{10} \). Give your answer correct to three significant figures.
[3 marks]
(c) Find the smallest value of \( n \) such that \( S_\infty – S_n < 0.5 \).
[4 marks]
Total: [9 marks]
What You Need to Know
📖 Key Information
Topic: AA SL 1.3 / AA HL 1.3 — Geometric sequences and series
Estimated time: 8–10 minutes
Calculator: Allowed (Paper 2 style)
You will need the following formulas from the IB Mathematics formula booklet:
Part (c) combines the sum to infinity with logarithmic reasoning — a favourite IB technique. If you need to brush up on how exponential and logarithmic thinking connects to series, check out our post on exponential functions practicecoming out next week.
HL Extension: Part (c) essentially asks you to explore the convergence behaviour of a geometric series. Understanding why the infinite sum exists when \( |r| < 1 \) — and how quickly the partial sums approach it — is a key conceptual point for HL students.
Hints
Hint 1 — Getting Started
For part (a), substitute directly into the general term formula \( u_n = u_1 \cdot r^{n-1} \) with \( n = 5 \).
(Try the question before reading further.)
Hint 2 — Part (b)
Use the finite geometric sum formula. Since \( r \neq 1 \), you can apply \( S_n = \frac{u_1(1 – r^n)}{1 – r} \) directly with \( n = 10 \). Keep exact values as long as possible before rounding.
(Have a go at part (c) now.)
Hint 3 — Part (c)
Write \( S_\infty – S_n \) as a single expression. You should find it simplifies to \( S_\infty \cdot r^n \). Then set up an inequality and use logarithms. Remember: when you divide by \( \ln\left(\frac{3}{4}\right) \), the inequality flips because \( \ln\left(\frac{3}{4}\right) < 0 \).
(Now try to finish the solution fully.)
Full Worked Solution
✍️ Step-by-Step Solution
Part (a): Find u₅
Apply the general term formula:
Part (b): Find S₁₀
Apply the finite sum formula with \( u_1 = 48 \), \( r = \frac{3}{4} \), \( n = 10 \):
Calculate \( \left(\frac{3}{4}\right)^{10} = \frac{3^{10}}{4^{10}} = \frac{59049}{1048576} \approx 0.05631\dots \)
Part (c): Find smallest n such that S∞ − Sₙ < 0.5
Step 1: First, confirm the series converges. Since \( |r| = \frac{3}{4} < 1 \), the sum to infinity exists.
Step 2: This is a key step in IB math sequences series practice — express the difference \( S_\infty – S_n \):
Step 3: Set up the inequality:
Step 4: Take natural logarithms of both sides:
Since \( \ln\left(\frac{3}{4}\right) < 0 \), dividing by it reverses the inequality:
Step 5: Since \( n \) must be a positive integer:
Examiner Notes
🎓 What the Examiner Wants to See
- Correct formula selection: The examiner checks that you choose the right formula (general term vs. sum) and substitute correctly. Write the formula first, then substitute — this earns method marks even if arithmetic goes wrong.
- Exact values where possible: In part (a), giving \( \frac{243}{16} \) is preferred. In part (b), the question specifies 3 significant figures, so follow that precisely.
- Convergence justification: In part (c), state why the infinite sum exists (\( |r| < 1 \)). This is especially important for HL candidates.
- Inequality reversal with logarithms: The examiner specifically watches for whether you flip the inequality when dividing by a negative number. Missing this loses the M1.
- Integer answer: The final answer must be a whole number. Writing \( n = 20.68 \) without rounding up to 21 loses the final A1.
Common Mistakes
❌ Common Mistakes
1. Wrong exponent in the general term
Many students write \( u_5 = 48 \cdot \left(\frac{3}{4}\right)^5 \) instead of \( 48 \cdot \left(\frac{3}{4}\right)^4 \). The formula is \( u_n = u_1 \cdot r^{n-1} \), so the exponent is always one less than the term number. This error costs you the A1 in part (a) and can cascade into later parts.
2. Forgetting to reverse the inequality
When you divide both sides of an inequality by \( \ln\left(\frac{3}{4}\right) \), which is negative, the direction of the inequality must flip. Forgetting this gives you \( n < 20.68 \) and a wrong final answer. The examiner awards zero for the method mark if the reversal is missing.
3. Not rounding up to the next integer
Part (c) asks for the smallest integer \( n \). Writing \( n = 20.68 \) or rounding down to \( n = 20 \) loses the final mark. Always round up when the question asks for the smallest \( n \) satisfying an inequality.
📚 Want More Practice?
Functions Workbook
Sequences and series practice covering arithmetic, geometric, and sigma notation — plus convergence and sum to infinity problems. Perfect if you want structured drills on exactly this type of question.
$9.55
If you solved all three parts correctly — especially the logarithmic inequality in part (c) — you’re in great shape for exam day. Keep building your skills with more IB math sequences series practice every week, and those marks will add up just like a convergent geometric series.
