Multiply again and again

The idea

A GP with first term a and common ratio r is a, ar, ar², ar³, …, and its nth term is tₙ = arⁿ⁻¹.

How many doublings?

The green squares are 3, 6, 12, 24, …: each stage doubles the stage before.

To reach the 6th stage from the 1st, how many times do we multiply by 2?

• 5 times
• 6 times
• 3 times

Stages of the pattern

Move through the stages. The orange squares are the ones that were just added by doubling.

The nth term

A GP with first term a and common ratio r is a, ar, ar², ar³, …, and its nth term is tₙ = arⁿ⁻¹.

For 3, 6, 12, 24, … : t₁ = 3, t₂ = 3 × 2, t₃ = 3 × 2², t₄ = 3 × 2³, so tₙ = 3 × 2ⁿ⁻¹. As a recursive rule: t₁ = 3 and tₙ = 2tₙ₋₁ for n ≥ 2.

Notes

A GP with first term a and common ratio r is a, ar, ar², ar³, …, and its nth term is tₙ = arⁿ⁻¹.

Check yourself

A GP has a = 3 and r = 2. What is t₅?

Answer: 48

t₅ = 3 × 2⁴ = 3 × 16 = 48.

A GP has a = 2 and r = 3. What is t₄?

Answer: 54

t₄ = 2 × 3³ = 2 × 27 = 54.

In tₙ = a × rⁿ⁻¹ why is the power n − 1 and not n?

The GP 1, 3, 9, 27, 81, … has a = 1 and r = 3. What is t₇?

Answer: 729

t₇ = 1 × 3⁶ = 729. Check: 81 × 3 = 243 (t₆), 243 × 3 = 729.

• Because r is always one less than n. r is the common ratio and has nothing to do with n. The power counts the multiplications.
• The first term a is not multiplied by r yet, so the nth term has n − 1 multiplications — correct. Yes. t₁ = a × r⁰ = a, and every later term multiplies by r one more time.
• It makes no difference. It does. With the power n, the first term would already be a × r instead of a.