First-Order Linear Recurrence Relations (Grade 12)
Free printable Grade 12 General Mathematics worksheet: the general first-order linear recurrence a(n+1) = r x a(n) + d, its nth term (iteration and closed form), and long-run steady-state behaviour (ACMGM075-077).
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Grade 12 · Math worksheet
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Math
Growth & Decay: First-Order Linear Recurrence Relations
Use each first-order linear recurrence a(n+1) = r x a(n) + d. Generate terms, find the nth term (iteration matches the closed form), and decide the long-run behaviour, including the steady-state value L = d/(1 - r) when 0 < r < 1.
- 1.A sequence is defined by a(1) = 6 and the first-order linear recurrence relation a(n+1) = 3 x a(n) + 3. List the first 4 terms.
- 2.A sequence is defined by a(1) = 1 and the first-order linear recurrence relation a(n+1) = 2 x a(n) + 5. List the first 4 terms.
- 3.A sequence is defined by a(1) = 5 and the first-order linear recurrence relation a(n+1) = 2 x a(n) + 2. List the first 4 terms.
- 4.A sequence is defined by a(1) = 5 and the first-order linear recurrence relation a(n+1) = 3 x a(n) + 3. List the first 4 terms.
- 5.A sequence follows a(1) = 6 and a(n+1) = 2 x a(n) + 7. Find the 6th term a(6).
- 6.A sequence follows a(1) = 2 and a(n+1) = 3 x a(n) + 6. Find the 6th term a(6).
- 7.A sequence follows a(1) = 4 and a(n+1) = 2 x a(n) + 3. Find the 4th term a(4).
- 8.A sequence follows a(1) = 27 and a(n+1) = 0.75 x a(n) + 7.5. Because the multiplier is between 0 and 1, the sequence approaches a long-run steady-state value. Find that steady-state value.
- 9.A sequence follows a(1) = 5 and a(n+1) = 2 x a(n) + 6. Does this sequence approach a steady-state value, or grow without bound? Answer "increasing" or "steady state".
- 10.A sequence follows a(1) = 34 and a(n+1) = 0.5 x a(n) + 19. Because the multiplier is between 0 and 1, the sequence approaches a long-run steady-state value. Find that steady-state value.
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