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Arithmetic & Geometric Series (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontSigma Notation
Define sigma notation.

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Cards in this collection (21)

Define sigma notation.
Sigma notation uses the capital Greek letter $Σ$ to mean 'the sum of'.
The expression to its right says what is being added, and the limits below and above say which terms to include.
What does $\sum_{k = 1}^{6} (2 \times 3^{k - 1})$ mean?
Work out $2 \times 3^{k - 1}$ for $k = 1$ , then $k = 2$ , and so on up to $k = 6$ , and add all six results.
It is the sum of the first six terms of that series, written compactly.
True or False?
The same sum can be written in sigma notation in more than one way.
True.
It can be left as $\sum_{k = 1}^{6} u_{k}$ or written out in full as $\sum_{k = 1}^{6} (2 \times 3^{k - 1})$ .
The counter letter is free as well, so $r$ would do the same job as $k$ .
For the series $u_{n} = 2 \times 3^{n - 1}$ , complete the two limits of the sum of the seventh to twelfth terms:
$\sum_{k =______}^{______} (2 \times 3^{k - 1})$
The completed sum is:
$\sum_{k = 7}^{12} (2 \times 3^{k - 1})$
The limits are term numbers, not the values of the terms, so they are simply $7$ and $12$ .
Why is the letter changed from $n$ to $k$ inside a sigma sum?
Because $k$ is the counter that runs through the limits, while $n$ is usually already being used for the number of terms.
Keeping them apart makes $\sum_{k = 1}^{n}$ readable: $k$ moves, and $n$ stays fixed.
How would you write the sum of the first $n$ terms of a geometric series in sigma notation?
As $\sum_{k = 1}^{n} a r^{k - 1}$ , putting the $n$ th term formula inside with $k$ as the counter.
An arithmetic series works the same way, as $\sum_{k = 1}^{n} (a + (k - 1) d)$ .
Define arithmetic sequence.
An arithmetic sequence has a common difference $d$ which is added to each term to get the next one.
The first term is called $a$ , so $1, 4, 7, 10, \dots$ has $a = 1$ and $d = 3$ .
What is the difference between a sequence and a series?
The terms are exactly the same in both; a series is what you get when those terms are added together.
So $1, 4, 7, 10$ is a sequence, while $1 + 4 + 7 + 10$ is the matching series.
True or False?
An arithmetic sequence must increase.
False.
A negative common difference makes the sequence decrease, term by term.
For instance $20, 17, 14, 11, \dots$ is arithmetic with $d = - 3$ .
Complete the $n$ th term formula for an arithmetic sequence:
$u_{n} = a + (n -______)______$
The completed formula is:
$u_{n} = a + (n - 1) d$
The $n - 1$ matters: the first term has had the common difference added to it no times yet.
What is the sum of the first $n$ terms of an arithmetic series?
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , where $a$ is the first term and $d$ the common difference.
The bracket is the first term plus the last, and $\frac{n}{2}$ counts how many such pairs there are.
Given the fourth and ninth terms of an arithmetic series, how do you find $a$ and $d$ ?
Put each into the $n$ th term formula to get two simultaneous equations.
With $u_{4} = 10$ and $u_{9} = 25$ that gives $a + 3 d = 10$ and $a + 8 d = 25$ , so $d = 3$ and $a = 1$ .
Define geometric sequence.
A geometric sequence has a common ratio $r$ which each term is multiplied by to get the next one.
The first term is called $a$ , so $2, 6, 18, 54, \dots$ has $a = 2$ and $r = 3$ .
How do you find the common ratio from two consecutive terms?
Divide a term by the one immediately before it, so $r = \frac{u_{n + 1}}{u_{n}}$ .
If the sixth term is $486$ and the seventh is $1458$ , then $r = \frac{1458}{486} = 3$ .
Complete the $n$ th term formula for a geometric sequence, filling in the missing index:
$u_{n} = a r^{______}$
The completed formula is:
$u_{n} = a r^{n - 1}$
The index is $n - 1$ rather than $n$ , because the first term has not been multiplied by $r$ at all.
What happens to a geometric sequence when $r$ is negative?
The terms alternate between positive and negative values.
Starting at $1$ with $r = - 4$ gives the sequence $1, - 4, 16, - 64, 256$ and so on.
What is the sum of the first $n$ terms of a geometric series?
$S_{n} = \frac{a (1 - r^{n})}{1 - r}$ , for any common ratio other than $1$ .
When $r > 1$ the rearrangement $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ avoids the negatives and gives the same answer.
With $a = 25$ and $r = 0.8$ , find the fifth term and the sum of the first five terms.
The fifth term is $25 \times 0 . 8^{4} = 10.24$ , using $n - 1 = 4$ as the index.
The sum is $S_{5} = \frac{25 (1 - 0 . 8^{5})}{1 - 0.8} = 84.04$ .
When does a geometric series converge?
When $|r| < 1$ , that is when $- 1 < r < 1$ , so the terms creep closer and closer to zero.
The running total then approaches a finite limiting value, called the sum to infinity.
What is the sum to infinity of a convergent geometric series?
$S_{\infty} = \frac{a}{1 - r}$ , which may only be used when $|r| < 1$ .
For $6, 2, \frac{2}{3}, \dots$ the ratio is $\frac{1}{3}$ , so $S_{\infty} = \frac{6}{1 - \frac{1}{3}} = 9$ .
True or False?
A geometric series with $r = - \frac{1}{2}$ has a sum to infinity.
True.
The condition is on the size of $r$ , not its sign, and $|- \frac{1}{2}|$ is $\frac{1}{2}$ , which is less than $1$ .
The terms alternate in sign but still shrink towards zero, so the running total settles on a finite value.

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