nth term of quadratic sequences Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE nth term of quadratic sequences questions. See exactly how to solve problems on second differences, quadratic sequences, coefficient of n squared, fractional coefficient.

second differencesquadratic sequencescoefficient of n squaredfractional coefficientsubstitutionquadratic nth term
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
Here are the first five terms of a quadratic sequence. 3, 7, 13, 21, 31, …3,\ 7,\ 13,\ 21,\ 31,\ \ldots Work out the second difference of the sequence.

Worked solution

  1. Work out the first differences

    4, 6, 8, 104,\ 6,\ 8,\ 10

    Take each term away from the term after it.

  2. Work out the second differences

    2, 2, 22,\ 2,\ 2

    Now difference the first differences.

  3. State the second difference

    22

    The second difference is constant, which confirms the sequence is quadratic.

Answer
22
Question 2
1 markeasy
Here are the first five terms of a quadratic sequence. 2, 10, 24, 44, 70, …2,\ 10,\ 24,\ 44,\ 70,\ \ldots Work out the second difference of the sequence.

Worked solution

  1. Work out the first differences

    8, 14, 20, 268,\ 14,\ 20,\ 26

    Subtract each term from the next one.

  2. Work out the second differences

    6, 6, 66,\ 6,\ 6

    Difference the first differences.

  3. State the second difference

    66

    A constant second difference of 66 tells you the n2n^2 coefficient will be 33.

Answer
66
Question 3
1 markeasy
The second difference of a quadratic sequence is 88. Write down the coefficient of n2n^2 in the nnth term of the sequence.

Worked solution

  1. Recall the link between a and the second difference

    second difference=2a\text{second difference} = 2a

    For the sequence an2+bn+can^2 + bn + c the second difference is always 2a2a.

  2. Halve the second difference

    a=82=4a = \frac{8}{2} = 4

    Halving is essential: 88 itself is 2a2a, not aa.

  3. State the coefficient

    44

    The nnth term is 4n2+bn+c4n^2 + bn + c for some bb and cc.

Answer
44
Question 4
2 markseasy
The second difference of a quadratic sequence is 33. Write down the coefficient of n2n^2 in the nnth term of the sequence.

Worked solution

  1. Recall the link between a and the second difference

    second difference=2a\text{second difference} = 2a

    The second difference of an2+bn+can^2+bn+c is 2a2a, whatever bb and cc are.

  2. Halve the second difference

    a=32a = \frac{3}{2}

    An odd second difference gives a fractional aa. Keep it as an exact fraction; do not round it.

  3. State the coefficient

    32\frac{3}{2}

    The nnth term is 32n2+bn+c\frac{3}{2}n^2 + bn + c.

Answer
32\frac{3}{2}
Question 5
1 markeasy
The nnth term of a quadratic sequence is 2n2+12n^2 + 1. Work out the 33rd term of the sequence.

Worked solution

  1. Substitute n equals 3

    2×32+12 \times 3^2 + 1

    Replace every nn in the rule with 33.

  2. Work out the square first

    2×9+1=18+12 \times 9 + 1 = 18 + 1

    Priority of operations: square before you multiply.

  3. State the term

    1919

    The 3rd term of the sequence is 1919.

Answer
1919

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