Here are the first four terms of a quadratic sequence - OCR - GCSE Maths - Question 15 - 2021 - Paper 1

Next, we find the second differences by subtracting consecutive first differences:

1. Second difference: 10 - 4 = 6
2. Third difference: 16 - 10 = 6

The second differences are constant at 6, indicating the quadratic nature of the sequence.

Since the second difference is 2a, we can set up the equation:

$2a = 6 \Rightarrow a = 3$.

Using the general form of the quadratic sequence, we can express the first few terms as:

1. For n = 1: (3(1)^2 + b(1) + c = -1 \Rightarrow 3 + b + c = -1 \Rightarrow b + c = -4)
2. For n = 2: (3(2)^2 + b(2) + c = 3 \Rightarrow 12 + 2b + c = 3 \Rightarrow 2b + c = -9)
3. For n = 3: (3(3)^2 + b(3) + c = 13 \Rightarrow 27 + 3b + c = 13 \Rightarrow 3b + c = -14)

Thus, we have a system of equations:

1. (b + c = -4)
2. (2b + c = -9)
3. (3b + c = -14).

From the equations, we can eliminate c:

1. Subtracting the first from the second: (2b + c - (b + c) = -9 - (-4) \Rightarrow b = -5)

2. Substituting (b = -5) into (b + c = -4): (-5 + c = -4 \Rightarrow c = 1)

Thus, we have:

• (a = 3)
• (b = -5)
• (c = 1).