Unlock Sequences and nth Term: Your Guide to Exam Success

What the examiner is testing

The examiner is assessing your ability to identify patterns in sequences and express these patterns algebraically using the \(n\)th term. Marks are awarded for correctly determining the \(n\)th term formula for linear and quadratic sequences, and for using this formula to find specific terms or to check if a number is part of a sequence.

The method

1. Write down the sequence and find the differences: List the given terms of the sequence. Calculate the difference between consecutive terms.
2. Check for linearity: If the first differences are constant, the sequence is linear. The constant difference is the coefficient of \(n\) in the \(n\)th term formula.
3. Find the "zeroth" term for linear sequences: Work backwards from the first term using the constant difference to find the term before the first term (the "zeroth" term). This is the constant term in the \(n\)th term formula.
4. Formulate the \(n\)th term for linear sequences: Combine the coefficient of \(n\) (the constant difference) and the "zeroth" term. For example, if the difference is \(d\) and the zeroth term is \(c\), the \(n\)th term is \(dn + c\).
5. Check for quadratic sequences: If the first differences are not constant but the second differences are constant, the sequence is quadratic. The coefficient of \(n^2\) is half of the constant second difference.
6. Find the \(n\)th term for quadratic sequences: Let the coefficient of \(n^2\) be \(a\). Subtract \(an^2\) from the original sequence terms to form a new sequence. This new sequence will be linear. Find the \(n\)th term of this new linear sequence using steps 2-4. Combine \(an^2\) with the \(n\)th term of the new linear sequence to get the full quadratic \(n\)th term.

Worked example

Find the \(n\)th term of the sequence \(5, 8, 11, 14, \dots\)

1. Write down the sequence and find the differences:
   $$5, \quad 8, \quad 11, \quad 14, \dots$$
   Differences:
   $$8 - 5 = 3$$
   $$11 - 8 = 3$$
   $$14 - 11 = 3$$
2. Check for linearity: The first differences are constant (\(3\)). So, it's a linear sequence. The coefficient of \(n\) is \(3\).
3. Find the "zeroth" term: Work backwards from the first term. The term before \(5\) would be \(5 - 3 = 2\).
4. Formulate the \(n\)th term: The \(n\)th term is \(3n + 2\).
5. Sanity check:
   For \(n=1\): \(3(1) + 2 = 5\) (Correct)
   For \(n=2\): \(3(2) + 2 = 8\) (Correct)
   For \(n=3\): \(3(3) + 2 = 11\) (Correct)

Worked example: a harder one

a) Find the \(n\)th term of the sequence \(4, 11, 22, 37, 56, \dots\)
b) Determine if \(199\) is a term in this sequence.

a) Find the \(n\)th term:
1. Write down the sequence and find the differences:
$$4, \quad 11, \quad 22, \quad 37, \quad 56, \dots$$
First differences:
$$11 - 4 = 7$$
$$22 - 11 = 11$$
$$37 - 22 = 15$$
$$56 - 37 = 19$$
The first differences (\(7, 11, 15, 19\)) are not constant, so it's not linear.
Second differences:
$$11 - 7 = 4$$
$$15 - 11 = 4$$
$$19 - 15 = 4$$
The second differences are constant (\(4\)), so it's a quadratic sequence.
2. The coefficient of \(n^2\) is half of the constant second difference: \(a = \frac{4}{2} = 2\). So the \(n\)th term starts with \(2n^2\).
3. Subtract \(2n^2\) from the original sequence:
Original sequence: \(4, 11, 22, 37, 56\)
\(2n^2\) sequence:
For \(n=1\): \(2(1)^2 = 2\)
For \(n=2\): \(2(2)^2 = 8\)
For \(n=3\): \(2(3)^2 = 18\)
For \(n=4\): \(2(4)^2 = 32\)
For \(n=5\): \(2(5)^2 = 50\)
New sequence (original term \(-\) \(2n^2\)):
\(4 - 2 = 2\)
\(11 - 8 = 3\)
\(22 - 18 = 4\)
\(37 - 32 = 5\)
\(56 - 50 = 6\)
The new sequence is \(2, 3, 4, 5, 6, \dots\).
4. Find the \(n\)th term of this new linear sequence:
Differences: \(1\).
Zeroth term: \(2 - 1 = 1\).
The \(n\)th term for this linear sequence is \(1n + 1\), or simply \(n+1\).
5. Combine to find the full \(n\)th term: \(2n^2 + n + 1\).
Sanity check:
For \(n=1\): \(2(1)^2 + 1 + 1 = 2+1+1 = 4\) (Correct)
For \(n=2\): \(2(2)^2 + 2 + 1 = 8+2+1 = 11\) (Correct)

b) Determine if \(199\) is a term in this sequence.
We need to solve \(2n^2 + n + 1 = 199\).
$$2n^2 + n + 1 = 199$$
$$2n^2 + n - 198 = 0$$
To find if \(n\) is an integer, we can use the quadratic formula:
$$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here, \(a=2\), \(b=1\), \(c=-198\).
$$n = \frac{-1 \pm \sqrt{1^2 - 4(2)(-198)}}{2(2)}$$
$$n = \frac{-1 \pm \sqrt{1 + 1584}}{4}$$
$$n = \frac{-1 \pm \sqrt{1585}}{4}$$
Calculate \(\sqrt{1585}\): \(\sqrt{1585} \approx 39.81\).
Since \(\sqrt{1585}\) is not an integer, \(n\) will not be an integer.
Therefore, \(199\) is not a term in the sequence.
Common mistake: Students might try to solve this by trial and error, which is inefficient and prone to error if the number is not a term. Setting up the equation and using the quadratic formula (or checking the discriminant) is the robust method.

Practice

1. Find the \(n\)th term of the sequence \(10, 13, 16, 19, \dots\)
2. The \(n\)th term of a sequence is \(5n - 2\). Write down the first three terms of the sequence.
3. A sequence has \(n\)th term \(n^2 + 3n\). What is the 5th term of the sequence?
4. Find the \(n\)th term of the sequence \(3, 8, 15, 24, 35, \dots\)

Answers:
1. \(3n + 7\)
2. \(3, 8, 13\)
3. \(40\)
Working for 4:
Sequence: \(3, 8, 15, 24, 35\)
First differences: \(5, 7, 9, 11\)
Second differences: \(2, 2, 2\)
Since the second difference is \(2\), the coefficient of \(n^2\) is \(\frac{2}{2} = 1\). So the \(n\)th term starts with \(n^2\).
Subtract \(n^2\) from the original sequence:
\(n^2\) sequence: \(1, 4, 9, 16, 25\)
New sequence (original term \(-\) \(n^2\)):
\(3 - 1 = 2\)
\(8 - 4 = 4\)
\(15 - 9 = 6\)
\(24 - 16 = 8\)
\(35 - 25 = 10\)
The new sequence is \(2, 4, 6, 8, 10, \dots\).
This is a linear sequence with a difference of \(2\). The zeroth term is \(2 - 2 = 0\).
The \(n\)th term for this linear sequence is \(2n + 0\), or \(2n\).
Combine: \(n^2 + 2n\).

The three mistakes that lose marks

1. Incorrectly calculating differences: A common error is to subtract in the wrong order or make arithmetic mistakes. For instance, in \(5, 8, 11, 14\), writing \(5-8 = -3\) instead of \(8-5 = 3\). This leads to an incorrect coefficient for \(n\).
2. Forgetting the "zeroth" term for linear sequences: Students often correctly identify the coefficient of \(n\) but then use the first term as the constant. For example, for \(5, 8, 11, 14\), finding the difference is \(3\), but then giving the \(n\)th term as \(3n+5\). This is incorrect because \(3(1)+5 = 8\), not \(5\). The correct "zeroth" term is \(2\), giving \(3n+2\).
3. Mixing up \(n^2\) and \(n\) terms in quadratic sequences: After finding the \(an^2\) part, students sometimes struggle to correctly find the \(bn+c\) part. Forgetting to subtract \(an^2\) from all terms of the original sequence before finding the linear part, or making arithmetic errors during this subtraction, leads to an incorrect final \(n\)th term. For example, if the sequence is \(4, 11, 22, \dots\) and \(2n^2\) is found, some might incorrectly try to find the \(bn+c\) from \(4, 11, 22\) directly instead of from \(2, 3, 4\).

30-second recap

A linear sequence has a constant first difference, and its \(n\)th term is \(dn+c\). A quadratic sequence has a constant second difference, and its \(n\)th term is \(an^2+bn+c\). To find \(a\) for quadratic sequences, halve the constant second difference.