IB Math AA SL Topic 1.3: Geometric sequences & series | Notes & Flashcards | Aimnova

IB Math AA SL — Geometric sequences & series

Topic 1.3 of IB Mathematics: Analysis and Approaches covers Geometric sequences & series, which is part of Unit 1: Number & Algebra. Students explore key concepts including nth term, Sum of n terms, Growth & decay. A strong understanding of geometric sequences & series is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Geometric sequences & series

Key Idea: A geometric sequence multiplies by the same ratio r each step — the engine behind compound interest, populations and bouncing balls. It runs through both papers: by-hand on Paper 1, GDC table / finance solver on Paper 2.

🔢 The two formulas

u_n = u_1 r^{\,n-1}

$u_1$: first term
$r$: common ratio — what you × by each step
$n$: term number (you ×r only n−1 times)

S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}, \quad r \neq 1

$S_n$: sum of the first n terms
$r > 1$: use (rⁿ − 1)/(r − 1)
$0 < r < 1$: use (1 − rⁿ)/(1 − r)

Geometric multiplies (× r); arithmetic adds (+ d). Divide consecutive terms — same ratio ⇒ geometric. 3, 6, 12, 24 (× 2) is geometric; 3, 6, 9, 12 (+ 3) is arithmetic.

✏️ IB-style worked examples

IB-style question — find the nth term

A geometric sequence has u₃ = 18 and u₆ = 486. Find an expression for the nth term uₙ.

Step by step:

Find r: divide the later value by the earlier, then take the (steps)-th root. The gap 6 − 3 = 3 steps.
$r = \sqrt[3]{\tfrac{486}{18}} = \sqrt[3]{27} = 3$
Step back to the first term: u₃ = u₁r², so divide by r².
$u_1 = \frac{18}{3^2} = 2$
Put u₁ = 2 and r = 3 into the formula.
$u_n = 2 \times 3^{\,n-1}$

Final answer:

uₙ = 2 × 3ⁿ⁻¹.

IB-style question — show a closed form for the sum (Paper 1)

A geometric sequence has u₁ = 4 and r = 3. Show that the sum of the first n terms is Sₙ = 2(3ⁿ − 1).

Step by step:

Write the sum formula.
$S_n = \frac{u_1(r^n - 1)}{r - 1}$
Substitute u₁ = 4 and r = 3.
$S_n = \frac{4(3^n - 1)}{3 - 1} = \frac{4(3^n - 1)}{2}$
Cancel to reach the required form.
$S_n = 2(3^n - 1)$

Final answer:

Sₙ = 2(3ⁿ − 1), as required.

IB-style question — depreciation by a fixed % (decay)

A machine worth $20 000 loses 15% of its value each year. After how many whole years is it first worth less than $8000?

Step by step:

Decay, so the ratio is 1 − the rate.
$r = 1 - 0.15 = 0.85$
Value after n years = start × rⁿ. Set it below 8000.
$20000 \times 0.85^{\,n} < 8000 \;\Rightarrow\; 0.85^{\,n} < 0.4$
Solve (logs or GDC) and round n up.
$n = 6$

Final answer:

After 6 years.

Important: You ×r one fewer time than the term number. 2016 → 2025 lists 10 years but only 9 increases happen; the 5th bounce is the 6th term (5 jumps, the drop is term 1). Sketch one arrow per step and count to the term you want.

Tap each card to reveal the answer.

Exam Tips

uₙ = u₁rⁿ⁻¹ — multiply by r only (n − 1) times; the first term is already there.
Two terms → r = the (steps)-th root of the value-ratio; an even gap means r could be ±.
Three terms geometric ⇒ middle² = product; if that gives a quadratic, report both values.
Sₙ needs u₁, r and n — pick the form that keeps top and bottom positive; "show" questions: simplify until it matches.
Context: r = 1 ± rate, value = A₀rⁿ; for "how long until" round n up and use the GDC table or TVM solver.

IB Math AA SL — Geometric sequences & series

Key concepts in Geometric sequences & series

Key Idea: A geometric sequence multiplies by the same ratio r each step — the engine behind compound interest, populations and bouncing balls. It runs through both papers: by-hand on Paper 1, GDC table / finance solver on Paper 2.

🔢 The two formulas

u_n = u_1 r^{\,n-1}

$u_1$: first term
$r$: common ratio — what you × by each step
$n$: term number (you ×r only n−1 times)

S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}, \quad r \neq 1

$S_n$: sum of the first n terms
$r > 1$: use (rⁿ − 1)/(r − 1)
$0 < r < 1$: use (1 − rⁿ)/(1 − r)

Geometric multiplies (× r); arithmetic adds (+ d). Divide consecutive terms — same ratio ⇒ geometric. 3, 6, 12, 24 (× 2) is geometric; 3, 6, 9, 12 (+ 3) is arithmetic.

✏️ IB-style worked examples

IB-style question — find the nth term

A geometric sequence has u₃ = 18 and u₆ = 486. Find an expression for the nth term uₙ.

Step by step:

Find r: divide the later value by the earlier, then take the (steps)-th root. The gap 6 − 3 = 3 steps.
$r = \sqrt[3]{\tfrac{486}{18}} = \sqrt[3]{27} = 3$
Step back to the first term: u₃ = u₁r², so divide by r².
$u_1 = \frac{18}{3^2} = 2$
Put u₁ = 2 and r = 3 into the formula.
$u_n = 2 \times 3^{\,n-1}$

Final answer:

uₙ = 2 × 3ⁿ⁻¹.

IB-style question — show a closed form for the sum (Paper 1)

A geometric sequence has u₁ = 4 and r = 3. Show that the sum of the first n terms is Sₙ = 2(3ⁿ − 1).

Step by step:

Write the sum formula.
$S_n = \frac{u_1(r^n - 1)}{r - 1}$
Substitute u₁ = 4 and r = 3.
$S_n = \frac{4(3^n - 1)}{3 - 1} = \frac{4(3^n - 1)}{2}$
Cancel to reach the required form.
$S_n = 2(3^n - 1)$

Final answer:

Sₙ = 2(3ⁿ − 1), as required.

IB-style question — depreciation by a fixed % (decay)

A machine worth $20 000 loses 15% of its value each year. After how many whole years is it first worth less than $8000?

Step by step:

Decay, so the ratio is 1 − the rate.
$r = 1 - 0.15 = 0.85$
Value after n years = start × rⁿ. Set it below 8000.
$20000 \times 0.85^{\,n} < 8000 \;\Rightarrow\; 0.85^{\,n} < 0.4$
Solve (logs or GDC) and round n up.
$n = 6$

Final answer:

After 6 years.

Important: You ×r one fewer time than the term number. 2016 → 2025 lists 10 years but only 9 increases happen; the 5th bounce is the 6th term (5 jumps, the drop is term 1). Sketch one arrow per step and count to the term you want.

Tap each card to reveal the answer.

Exam Tips

uₙ = u₁rⁿ⁻¹ — multiply by r only (n − 1) times; the first term is already there.
Two terms → r = the (steps)-th root of the value-ratio; an even gap means r could be ±.
Three terms geometric ⇒ middle² = product; if that gives a quadratic, report both values.
Sₙ needs u₁, r and n — pick the form that keeps top and bottom positive; "show" questions: simplify until it matches.
Context: r = 1 ± rate, value = A₀rⁿ; for "how long until" round n up and use the GDC table or TVM solver.

IB Math AA SL — Geometric sequences & series

Key concepts in Geometric sequences & series

🔢 The two formulas

✏️ IB-style worked examples

IB-style question — find the nth term

IB-style question — show a closed form for the sum (Paper 1)

IB-style question — depreciation by a fixed % (decay)

Exam Tips

What you'll learn in Topic 1.3

Study resources — 1.3 Geometric sequences & series

nth term

Sum of n terms

Growth & decay

Ready to study Geometric sequences & series?

Ready to practice?

IB Math AA SL — Geometric sequences & series

Key concepts in Geometric sequences & series

🔢 The two formulas

✏️ IB-style worked examples

IB-style question — find the nth term

IB-style question — show a closed form for the sum (Paper 1)

IB-style question — depreciation by a fixed % (decay)

Exam Tips

What you'll learn in Topic 1.3

Study resources — 1.3 Geometric sequences & series

nth term

Sum of n terms

Growth & decay

Ready to study Geometric sequences & series?

Ready to practice?