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Arithmetic & Geometric Progressions (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontLanguage of Sequences & Series
Define a sequence and a term.

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Define a sequence and a term.
A sequence, also called a progression, is an ordered list of numbers together with a rule for generating them.
Each number in it is a term, and terms are usually labelled $u_{1}$ , $u_{2}$ , $u_{3}$ and so on.
What is the difference between a sequence and a series?
A sequence is the list of terms; a series is what you get by adding those terms together.
So $1, 3, 5, 7$ is a sequence, while $1 + 3 + 5 + 7$ is the matching series.
True or False?
$S_{n}$ means the $n$ th term of a sequence.
False.
$S_{n}$ is the sum of the first $n$ terms, so $S_{n} = u_{1} + u_{2} + \dots + u_{n}$ .
The $n$ th term on its own is written $u_{n}$ .
A sequence is defined by $u_{n} = 5 - 2 n$ . Find the first five terms and $S_{5}$ .
Substituting $n = 1$ to $5$ gives the terms $3$ , $1$ , $- 1$ , $- 3$ , $- 5$ .
Adding those together gives $S_{5} = - 5$ , which is negative because the later terms outweigh the early ones.
What does the symbol $Σ$ tell you to do?
$Σ$ is the Greek capital sigma and stands for sum.
The expression to its right says what is being added, and the numbers below and above it say where the counting starts and stops.
True or False?
The number below the $Σ$ is always 1.
False.
The lower limit can be any value, so $\sum_{k = 0}^{4} (2 k + 1)$ and $\sum_{k = 7}^{14} (2 k - 13)$ are both perfectly ordinary sums.
Always read the limits rather than assuming the count begins at 1.
Define an arithmetic progression.
An arithmetic progression is a sequence in which the difference between consecutive terms is constant.
That constant is called the common difference, $d$ , and the sequence starts at the first term, $a$ .
Complete the formula for the $n$ th term of an arithmetic progression:
$u_{n} =______$
The completed formula is:
$u_{n} = a + (n - 1) d$
Here $a$ is the first term and $d$ the common difference.
Why does the formula use $(n - 1) d$ rather than $n d$ ?
Because the first term has had no common differences added to it yet.
Check it with $n = 1$ : the formula gives $a + 0 \times d = a$ , which is right, whereas $n d$ would wrongly add one difference straight away.
True or False?
The common difference of an arithmetic progression can be negative.
True.
A negative common difference simply makes the sequence decrease, as in $20, 17, 14, 11, \dots$ where $d = - 3$ .
The word progression says nothing about the direction of travel.
Complete the two forms for the sum of the first $n$ terms:
$S_{n} = \frac{1}{2} n (a +______)$
$S_{n} = \frac{1}{2} n (2 a +______)$
The completed forms are:
$S_{n} = \frac{1}{2} n (a + l)$
$S_{n} = \frac{1}{2} n (2 a + (n - 1) d)$
Here $l$ is the last term, and the second form is just the first with $l$ replaced by $a + (n - 1) d$ .
Which form of the sum formula should you use?
Use $\frac{1}{2} n (a + l)$ when you know the first and last terms.
Use $\frac{1}{2} n (2 a + (n - 1) d)$ when you know the first term and the common difference but not the last term.
An arithmetic progression has first term 3 and twentieth term 60. Find $S_{20}$ .
Both the first and last terms are known, so use the shorter form:
$S_{20} = \frac{1}{2} \times 20 \times (3 + 60) = 630$
No common difference is needed, which is why this form is worth spotting.
A question gives you $S_{n}$ and the first term, and asks for the common difference. What do you do?
Substitute everything you know into $S_{n} = \frac{1}{2} n (2 a + (n - 1) d)$ and solve the resulting equation for $d$ .
The formulas work in any direction, so an unknown inside one is found the same way as the sum itself.
Define a geometric progression.
A geometric progression is a sequence in which consecutive terms are related by a constant multiplier.
That multiplier is the common ratio, $r$ , and the sequence starts at the first term, $a$ .
Complete the formula for the $n$ th term of a geometric progression:
$u_{n} =______$
The completed formula is:
$u_{n} = a r^{n - 1}$
The power is $n - 1$ for the same reason as in an arithmetic progression: the first term has not yet been multiplied by anything.
What happens to a geometric progression when $r$ is negative?
The terms alternate between positive and negative.
For instance $1, - 4, 16, - 64, 256, \dots$ has $r = - 4$ , and the signs flip with every multiplication.
You are given two consecutive terms of a geometric progression. How do you find $r$ ?
Divide a term by the one immediately before it, since that division undoes the multiplication.
With $r$ known, substituting either term into $u_{n} = a r^{n - 1}$ then gives the first term.
What is a geometric series?
It is the sum of the terms of a geometric progression.
So the progression $2, 6, 18, 54, \dots$ has the matching series $2 + 6 + 18 + 54 + \dots$
Complete the sum of the first $n$ terms of a geometric progression:
$S_{n} = \frac{a (1 -______)}{1 - r}$
The completed formula is:
$S_{n} = \frac{a (1 - r^{n})}{1 - r}$
There is an equivalent form, $\frac{a (r^{n} - 1)}{r - 1}$ , and both require $r \neq 1$ .
Which form of the geometric sum formula is more convenient?
Use $\frac{a (1 - r^{n})}{1 - r}$ when $r < 1$ , so that both the top and bottom stay positive.
Use $\frac{a (r^{n} - 1)}{r - 1}$ when $r > 1$ , for the same reason, since the two are just each other multiplied through by $- 1$ .
True or False?
Every geometric series has a sum to infinity.
False.
A sum to infinity exists only when $| r | < 1$ , and it is then $S_{\infty} = \frac{a}{1 - r}$ .
If $| r | \geq 1$ the terms do not shrink away, so the total grows without limit and the series diverges.

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