Define a linear sequence.

A linear sequence goes up or down by the same amount each time, so its first differences are constant. It is also called an arithmetic sequence.

How do you recognise a quadratic sequence?

Its first differences are not constant, but its second differences are. Another way to see it is that the sequence of first differences is itself a linear sequence.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsSequences

Sequences (AQA GCSE Further Maths): Flashcards

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FrontFinding nth Terms of Sequences
Define a linear sequence.

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Define a linear sequence.
A linear sequence goes up or down by the same amount each time, so its first differences are constant.
It is also called an arithmetic sequence.
How do you find the nth term of a linear sequence?
The form is $d n + b$ , where $d$ is the common difference between the terms.
Find $d$ first, then substitute $n = 1$ and the first term to work out $b$ .
Find the nth term of 5, 7, 9, 11, 13, …
The common difference is 2, so the nth term is $2 n + b$ .
Substituting $n = 1$ and the first term 5 gives $5 = 2 + b$ , so $b = 3$ and the nth term is $2 n + 3$ .
True or False?
A sequence going down by 2 each time has $2 n$ in its nth term.
False.
A decreasing sequence has $- d n$ , so going down by 2 gives $- 2 n$ .
For 5, 3, 1, $- 1$ , … the nth term is $- 2 n + 7$ .
How do you recognise a quadratic sequence?
Its first differences are not constant, but its second differences are.
Another way to see it is that the sequence of first differences is itself a linear sequence.
Complete the rule for a quadratic sequence with nth term $a n^{2} + b n + c$ :
$a$ is the second difference divided by $______$ .
The completed rule is:
$a$ is the second difference divided by 2.
So a second difference of 4 gives $a = 2$ , because the second difference is always twice the coefficient of $n^{2}$ .
Once you have $a$ , how do you find $b$ and $c$ ?
Write out $a n^{2}$ alongside the sequence and subtract, term by term.
Those differences form a linear sequence, whose nth term is $b n + c$ .
Find the nth term of 5, 7, 11, 17, 25, …
The second differences are 2, so $a = 1$ , and subtracting $n^{2}$ from the sequence leaves 4, 3, 2, 1, …
That linear sequence has nth term $- n + 5$ , so the answer is $n^{2} - n + 5$ .
How can you find the nth term of 4, 7, 12, 19, 28, … without using differences?
Compare it with the square numbers 1, 4, 9, 16, 25: every term is exactly 3 more.
So the nth term is $n^{2} + 3$ , with no need for the second-difference method at all.
What values can $n$ take in an nth term formula?
Only positive integers, since $n$ counts the position of a term in the sequence.
So substituting $n = 1$ gives the first term, $n = 2$ the second, and so on.
How do you find which term of $\frac{4 n}{n + 1}$ equals $\frac{18}{5}$ ?
Set the formula equal to $\frac{18}{5}$ and solve the equation for $n$ .
That gives $n = 9$ , so it is the 9th term.
True or False?
In an nth term formula, $n$ and the value of the term are the same thing.
False.
$n$ is the position in the sequence, while the value is whatever the formula gives for that position.
For $\frac{4 n}{n + 1}$ the 4th term has $n = 4$ but a value of $\frac{16}{5}$ .
How do you find the first negative term of $\frac{5 - n}{2 n^{2} + 1}$ ?
Find where the formula is zero: $5 - n = 0$ gives $n = 5$ , so the 5th term is zero.
Since the terms are decreasing, the 6th is the first negative one, and checking gives $- \frac{1}{73}$ .
Define the limiting value of a sequence.
The limiting value is the value the terms get closer and closer to as $n$ increases.
It is what the formula tends to as $n \to \infty$ , read as " $n$ tends to infinity".
Complete the limiting value as $n$ tends to infinity:
$\frac{1}{n} \to______$
The completed statement is:
$\frac{1}{n} \to 0$
The same holds for $\frac{10}{n}$ , $\frac{1}{n^{2}}$ and $- \frac{4}{n^{8}}$ : any constant over a power of $n$ tends to zero.
How do you find the limiting value of an algebraic fraction in $n$ ?
Divide every term, on the top and on the bottom, by the highest power of $n$ present.
Every term that becomes a constant over a power of $n$ then tends to zero, leaving the limit behind.
Find the limiting value of $\frac{5 n + 3}{2 n - 1}$ .
Dividing the top and bottom by $n$ gives $\frac{5 + \frac{3}{n}}{2 - \frac{1}{n}}$ .
Both of those fractions tend to zero, so the limiting value is $\frac{5}{2}$ .
True or False?
Some sequences have no limiting value at all.
True.
The sequence $5 n$ gives 5, 10, 15, 20, … and never settles towards any value.
A limiting value only exists where the terms close in on something.

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