Quadratic Sequences Worksheet 1 | PDF - Free Printable

Explanation:
We are given two sequences and asked to find a formula for the *n*th term of each. These are quadratic sequences, meaning the second differences are constant (not the first differences — that would be linear). Let’s solve each exercise step by step.

---

Exercise 1: Sequence: 6, 11, 18, 27, 38, 51, …

Step 1: Write the terms with their positions (*n* = 1, 2, 3, …):

| n | aₙ |
|---|----|
| 1 | 6 |
| 2 | 11 |
| 3 | 18 |
| 4 | 27 |
| 5 | 38 |
| 6 | 51 |

Step 2: Find first differences (Δ¹):
11 − 6 = 5
18 − 11 = 7
27 − 18 = 9
38 − 27 = 11
51 − 38 = 13

So first differences: 5, 7, 9, 11, 13

Step 3: Find second differences (Δ²):
7 − 5 = 2
9 − 7 = 2
11 − 9 = 2
13 − 11 = 2

Second differences are constant = 2 → confirms it's quadratic.

For a quadratic sequence, the general form is:
aₙ = An² + Bn + C
And the second difference = 2A ⇒ 2A = 2 ⇒ A = 1.

So now we have:
aₙ = n² + Bn + C

Now plug in values for n = 1 and n = 2 to solve for B and C.

For n = 1: a₁ = 6 = (1)² + B(1) + C = 1 + B + C
→ B + C = 5 …(i)

For n = 2: a₂ = 11 = (2)² + B(2) + C = 4 + 2B + C
→ 2B + C = 7 …(ii)

Subtract (i) from (ii):
(2B + C) − (B + C) = 7 − 5
→ B = 2

Plug back into (i):
2 + C = 5 → C = 3

So formula: aₙ = n² + 2n + 3

Let’s verify with n = 3:
3² + 2·3 + 3 = 9 + 6 + 3 = 18 ✔
n = 4: 16 + 8 + 3 = 27 ✔
n = 5: 25 + 10 + 3 = 38 ✔
Good.

---

Exercise 2: Sequence: −3, 0, 5, 12, 21, 32, …

Step 1: List terms with n:

| n | aₙ |
|---|----|
| 1 | −3 |
| 2 | 0 |
| 3 | 5 |
| 4 | 12 |
| 5 | 21 |
| 6 | 32 |

Step 2: First differences:
0 − (−3) = 3
5 − 0 = 5
12 − 5 = 7
21 − 12 = 9
32 − 21 = 11

First differences: 3, 5, 7, 9, 11

Step 3: Second differences:
5 − 3 = 2
7 − 5 = 2
9 − 7 = 2
11 − 9 = 2

Constant second difference = 2 ⇒ again, quadratic, and 2A = 2 ⇒ A = 1.

So aₙ = n² + Bn + C

Use n = 1: a₁ = −3 = 1² + B(1) + C = 1 + B + C
→ B + C = −4 …(i)

n = 2: a₂ = 0 = 4 + 2B + C
→ 2B + C = −4 …(ii)

Subtract (i) from (ii):
(2B + C) − (B + C) = −4 − (−4) = 0
→ B = 0

Then from (i): 0 + C = −4 ⇒ C = −4

So formula: aₙ = n² − 4

Check:
n = 1: 1 − 4 = −3 ✔
n = 2: 4 − 4 = 0 ✔
n = 3: 9 − 4 = 5 ✔
n = 4: 16 − 4 = 12 ✔
n = 5: 25 − 4 = 21 ✔
n = 6: 36 − 4 = 32 ✔
Perfect.

Final Answer:
Exercise 1: $ a_n = n^2 + 2n + 3 $
Exercise 2: $ a_n = n^2 - 4 $