Add d to the last one

The idea

An AP with first term a and common difference d has the recursive rule t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2.

From stage 5 to stage 6

Back to the squares: 1, 5, 9, 13, 17, … Every term is 4 more than the previous term.

Stage 5 has 17 squares. How would you get the number of squares at stage 6?

• Add 4 to 17
• Multiply 17 by 4
• Add 6 to 17

Write the rule

Set the first term a and the difference d. The rule builds terms, and you compare them with the target.

The rule for an AP

An AP with first term a and common difference d has the recursive rule t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2.

For 1, 5, 9, 13, 17, … the rule is t₁ = 1, tₙ = tₙ₋₁ + 4 for n ≥ 2. Every term is 4 more than the previous one.

Notes

An AP with first term a and common difference d has the recursive rule t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2.

Check yourself

An AP has t₁ = 7 and tₙ = tₙ₋₁ + 3 for n ≥ 2. What is t₄?

Answer: 16

t₂ = 10, t₃ = 13, t₄ = 16.

Which is the recursive rule of the AP 2, 9, 16, 23, …?

An AP has t₁ = 20 and tₙ = tₙ₋₁ − 6 for n ≥ 2. What is t₃?

Answer: 8

t₂ = 20 − 6 = 14, t₃ = 14 − 6 = 8.

Which recursive rule gives the AP 11, 7, 3, −1, −5, …?

• t₁ = 2, tₙ = tₙ₋₁ + 2. You used the first term as the difference. The difference is 9 − 2 = 7.
• t₁ = 7, tₙ = tₙ₋₁ + 2. The first term is 2 and the difference is 7. These two are swapped.
• t₁ = 2, tₙ = tₙ₋₁ + 7 — correct. Yes. The first term is 2, and each term is 7 more than the one before.
• t₁ = 11, tₙ = tₙ₋₁ + (−4) — correct. Yes. The common difference is −4, so each term is the previous one plus (−4).
• t₁ = 11, tₙ = tₙ₋₁ + 4. That would make the terms grow: 11, 15, 19, … The terms here go down by 4.
• t₁ = −4, tₙ = tₙ₋₁ + 11. The first term is 11 and the difference is −4. These two are swapped.