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Algebra - Quadratic sequences (nth term)

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A quadratic sequence is a sequence of numbers where the second difference between consecutive terms is constant.

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The general form of the nthn^{th} term for a quadratic sequence is Tn=an2+bn+cT_n = an^2 + bn + c, where a,b, and ca, b, \text{ and } c are constants and a≠0a \neq 0.

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To find the value of aa, use the relationship: 2a=second difference2a = \text{second difference}.

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To find the values of bb and cc, you can subtract the an2an^2 term from the original sequence to find a remaining linear sequence, or solve the system of equations: 3a+b=first difference of the first two terms3a + b = \text{first difference of the first two terms} and a+b+c=first terma + b + c = \text{first term}.

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If the second difference is positive, the sequence grows quadratically upwards; if negative, it grows downwards.

📐Formulae

Tn=an2+bn+cT_n = an^2 + bn + c

2a=second difference2a = \text{second difference}

3a+b=T2−T13a + b = T_2 - T_1

a+b+c=T1a + b + c = T_1

💡Examples

Problem 1:

Find the nthn^{th} term formula for the sequence: 6,13,22,33,46,…6, 13, 22, 33, 46, \dots

Solution:

  1. Find the first differences: 13−6=713-6=7, 22−13=922-13=9, 33−22=1133-22=11, 46−33=1346-33=13.
  2. Find the second differences: 9−7=29-7=2, 11−9=211-9=2, 13−11=213-11=2. The constant second difference is 22.
  3. Find aa: 2a=2  ⟹  a=12a = 2 \implies a = 1
  4. Find bb: 3a+b=7  ⟹  3(1)+b=7  ⟹  b=43a + b = 7 \implies 3(1) + b = 7 \implies b = 4
  5. Find cc: a+b+c=6  ⟹  1+4+c=6  ⟹  c=1a + b + c = 6 \implies 1 + 4 + c = 6 \implies c = 1
  6. The nthn^{th} term is Tn=n2+4n+1T_n = n^2 + 4n + 1.

Explanation:

We identify the second difference to determine the quadratic coefficient aa, then use the linear relationships derived from the first term and the first difference to solve for bb and cc.

Problem 2:

Determine the nthn^{th} term of the sequence: 2,9,20,35,54,…2, 9, 20, 35, 54, \dots

Solution:

  1. First differences: 7,11,15,197, 11, 15, 19.
  2. Second differences: 4,4,44, 4, 4. Constant difference is 44.
  3. a=42=2a = \frac{4}{2} = 2.
  4. Subtract an2an^2 (2n22n^2) from the sequence:
    • For n=1:2−2(1)2=0n=1: 2 - 2(1)^2 = 0
    • For n=2:9−2(2)2=1n=2: 9 - 2(2)^2 = 1
    • For n=3:20−2(3)2=2n=3: 20 - 2(3)^2 = 2
    • For n=4:35−2(4)2=3n=4: 35 - 2(4)^2 = 3
  5. The remaining sequence is 0,1,2,3,…0, 1, 2, 3, \dots which is a linear sequence with nthn^{th} term n−1n - 1.
  6. Combine the parts: Tn=2n2+n−1T_n = 2n^2 + n - 1.

Explanation:

This method involves removing the quadratic component an2an^2 and finding the nthn^{th} term of the resulting linear sequence (bn+cbn + c).