Here are the first five terms of a quadratic sequence - Edexcel - GCSE Maths - Question 17 - 2020 - Paper 2

Next, we find the second differences of the first differences.

Second differences:

• 17 - 11 = 6
• 23 - 17 = 6
• 29 - 23 = 6

The second differences are constant at 6, which indicates that the sequence is indeed quadratic.

Since the second differences are constant, we can express the nth term in the form:

$a_n = an^2 + bn + c$

Where a, b, and c are constants. Since the second difference is 6, we can determine that:

$a = \frac{6}{2} = 3$

Thus, the expression so far is:

$a_n = 3n^2 + bn + c$.

We can use the known terms of the sequence to find the constants b and c.

Using values for n:

• For n=1, a_1 = 10: $3(1^2) + b(1) + c = 10 \Rightarrow 3 + b + c = 10$
• For n=2, a_2 = 21: $3(2^2) + b(2) + c = 21 \Rightarrow 12 + 2b + c = 21$
• For n=3, a_3 = 38: $3(3^2) + b(3) + c = 38 \Rightarrow 27 + 3b + c = 38$

We now have a system of equations to solve for b and c.

From the equations:

1. $b + c = 7$ (1)
2. $2b + c = 9$ (2)
3. $3b + c = 11$ (3)

Subtracting equation (1) from (2): $2b + c - (b + c) = 9 - 7 \Rightarrow b = 2$

Now substituting b back into (1): $2 + c = 7 \Rightarrow c = 5$

Thus, we have: $a = 3, b = 2, c = 5$.

Putting it all together, the expression for the nth term of the sequence is:

$a_n = 3n^2 + 2n + 5$