Here are the first six terms of a quadratic sequence - Edexcel - GCSE Maths - Question 17 - 2019 - Paper 3

Now calculate the second differences by subtracting the first differences:

• Second differences: 10 - 6 = 4
• Third differences: 14 - 10 = 4
• Fourth differences: 18 - 14 = 4
• Fifth differences: 22 - 18 = 4

The second differences are constant and equal to 4.

Since the second differences are constant, we know that the nth term can be expressed as a quadratic polynomial:

$T_n = an^2 + bn + c$

To find the coefficients a, b, and c:

1. Since the second difference is 4, we have: $2a = 4 \implies a = 2$

2. Substituting known positions into the equation will help find b and c. Using the first term, n = 1: $T_1 = 2(1^2) + b(1) + c = -1 \implies 2 + b + c = -1$ Simplifying gives us: $b + c = -3$

3. Using the second term, n = 2: $T_2 = 2(2^2) + b(2) + c = 5 \implies 8 + 2b + c = 5$ Simplifying gives: $2b + c = -3$

4. Now, subtract these two equations:

   • From these equations, we can solve for b and c:
   • Solving gives b = -2 and c = -1

Thus, the nth term of the sequence is:

$T_n = 2n^2 - 2n - 1$