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Here are the first four terms of a quadratic sequence - OCR - GCSE Maths - Question 17 - 2018 - Paper 4

Question 17

Here are the first four terms of a quadratic sequence. 2 15 34 59 The nth term is an² + bn + c. Find the values of a, b and c.

Worked Solution & Example Answer:Here are the first four terms of a quadratic sequence - OCR - GCSE Maths - Question 17 - 2018 - Paper 4

Step 1

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Answer

To identify the coefficients of the quadratic sequence, we start by calculating the first and second differences of the given terms:

This gives us the first difference sequence: 13, 19, 25.

The second differences are constant (6), confirming that the sequence is quadratic.

Step 2

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Answer

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

$2a = 6$

Solving for 'a' gives us:

$a = 3$ .

Step 3

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Using the quadratic formula $n^2 + bn + c$ , we can establish the first term when n = 1:

$3(1^2) + b(1) + c = 2$ , which simplifies to:

$3 + b + c = 2\Rightarrow b + c = -1. ag{1}$

Now, for the second term (n = 2):

$3(2^2) + b(2) + c = 15$ , leading to:

$12 + 2b + c = 15\Rightarrow 2b + c = 3. ag{2}$

Now we solve equations (1) and (2):

From (1), we have: $c = -1 - b. ag{3}$ Substituting (3) into (2) gives: $2b + (-1 - b) = 3\Rightarrow b - 1 = 3\Rightarrow b = 4.$

Substituting b back into (3): $c = -1 - 4 = -5.$

Therefore, we have: