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Sequences - EduqasFind the nth term of quadratic sequences - Higher

Sequences can be linear, quadratic or practical and based on real-life situations. Finding general rules helps find terms in sequences.

Part of MathsAlgebra

Finding the nth term of quadratic sequences - Higher

Quadratic sequences are sequences that include an \(n^2\) term. They can be identified by the fact that the differences in-between the terms are not equal, but the second differences between terms are equal.

Example 1

Work out the nth term of the sequence 2, 5, 10, 17, 26, ...

Sequence showing that it's increments are equally incremental (+2).

Work out the first differences between the terms. The first differences are not the same, so work out the second differences.

The second differences are the same. The sequence is quadratic and will contain an \(n^2\) term. The of \(n^2\) is always half of the second difference. In this example, the second difference is 2. Half of 2 is 1, so the coefficient of \(n^2\) is 1.

To work out the nth term of the sequence, write out the numbers in the sequence \(n^2\) and compare this sequence with the sequence in the question.

\(n^2\)14916
Operation\(+ 1\)\(+ 1\)\(+ 1\)\(+ 1\)
Sequence251017
\(n^2\)
1
4
9
16
Operation
\(+ 1\)
\(+ 1\)
\(+ 1\)
\(+ 1\)
Sequence
2
5
10
17

In this example, you need to add \(1\) to \(n^2\) to match the sequence. The sequence is therefore \(n^2 + 1\).

Example 2

Work out the nth term of the sequence 5, 11, 21, 35, ...

Sequence showing that it's increments are equally incremental (+4)

Work out the first differences between the terms. The first differences are not the same, so work out the second differences.

The second difference is the same so the sequence is quadratic and will contain an \(n^2\) term. The coefficient of \(n^2\) is half the second difference, which is 2. The sequence will contain \(2n^2\), so use this:

\(2n^2\)281832
Operation\(+ 3\)\(+ 3\)\(+ 3\)\(+ 3\)
Sequence5112135
\(2n^2\)
2
8
18
32
Operation
\(+ 3\)
\(+ 3\)
\(+ 3\)
\(+ 3\)
Sequence
5
11
21
35

The sequence is \(2n^2 + 3\).

Geometric sequences - Higher

In a geometric sequence, the term to term rule is to multiply or divide by the same value. This value is called the common ratio, \(r\), which can be worked out by dividing one term by the previous term.

Example 1

Show that the sequence 3, 6, 12, 24, 鈥 is a geometric sequence, and find the next three terms.

Dividing each term by the previous term gives the same value: \(6 \div 3 = 12 \div 6 = 24 \div 12 = 2\). So the common ratio is 2 and this is therefore a geometric sequence.

The next three terms are: \(24 \times 2 = 48\), \(48 \times 2 = 96\) and \(96 \times 2 = 192\).

Example 2

Find the next three terms in the geometric sequences:

a) 6, 4.2, 2.94, ...

b) 3, \(3\sqrt{3}\), 9, \(9\sqrt{3}\), 27, ...

c) 2, -4, 8, -16,...

a) To find the value of the common ratio, work out \( 4.2 \div 6 = 0.7\). The next three terms of the sequence are \(2.94 \times 0.7 = 2.058\), \(2.058 \times 0.7 = 1.4406\) and \(1.4406 \times 0.7 =1.00842\). The common ratio is less than 1. If the size of the common ratio is less than 1, the terms of the sequence will reduce in size.

b) The common ratio is \(3\sqrt{3} \div3 = \sqrt{3}\). The next three terms are \(27 \times \sqrt{3} = 27\sqrt{3}\), \(27\sqrt{3} \times \sqrt{3} = 27 \times 3 = 81,\) and \(81 \times \sqrt{3} = 81\sqrt{3}\). Some of the terms of this sequence are surds, so leave your answer in surds as this is more accurate than writing them in decimal form as they would have to be rounded.

c) The common ratio is \(-4 \div 2 = -2\). The next three terms of the sequence are \(-16 \times -2 = 32\), \(32 \times -2 = -64\), and \(-64 \times -2 = 128\). The common ratio is negative which means that the terms of the sequence will alternate between positive and negative.

Finding the nth term of a geometric sequence

The nth term of a geometric sequence is \(ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio.

Example

Find the nth term of the geometric sequence: 2, 2.4, 2.88, 3.456 and then find the 10th term.

The first term is 2, so \(a = 2\).

The common ratio is \(2.4 \div 2 = 1.2\), so \(r = 1.2\).

The nth term of the geometric sequence is \(ar^{n-1} = 2 \times 1.2^{n-1}\)

The 10th term is \(2 \times 1.2^9= 10.3195607\)