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Transcript
 Sequences A sequence is a list of numbers, or a pattern, which obeys a rule.
Each number in a sequence is called a term. ie the fourth term of the sequence 2, 4, 6, 8, 10, 12.... is 8, because it is the fourth number in the sequence.
With sequences if we are given the rule we can make the sequence or, eventually, as you improve you can find the rule when given the sequence.
Your knowledge of sequences begins with simple things like counting up in tens to recognising the sequence of odd or even numbers
Once you have this grasped then it is about recognising sequences that go up in equal steps such as 2, 5 or 10 and being able to extend them forwards and backwards.
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e.g write down the next 4 terms in the sequence 14, 19, 24, 29.....
Look for the difference between each term
This sequence is increasing by 5 each term so we continue it in this way 4 more times to get.....
14, 19, 24, 29, 34, 39, 44, 49
Sometimes you have to go backwards in a sequence. When doing this we do the inverse (opposite) from what we do going the normal way.
Knowing this I can work out the 2 terms that come before the 14 in the sequence above.
Since we were adding 5 going the normal way, we do the inverse going backwards so we subtract 5 each time to get.....
4, 9, 14, 19, 24, 29, 34, 39, 44, 49
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Sometimes you have a sequence with decimal numbers that you have to extend forwards or backwards
The method for this is the same as a whole number sequence in that you need to find the difference between each term. This will require a good grasp of place value as it involves decimals
e.g Calculate the 3 terms before and the 3 terms after ..... 5.5, 6.2, 6.9, 7.6 .....
The difference between 6.2 and 5.5 is 0.7 so this means that we add 0.7 each time going to the right and subtract 0.7 each time when going to the left.
This means that we end up with.....
3.4, 4.1, 4.8, 5.5, 6.2, 6.9, 7.6, 8.3, 9.0, 9.7
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The next step in sequences is where you are given the rule to the sequence (in words) and you have to make the sequence. e.g write the first five terms of the sequence 'double and add 3'
The rule is key:
To find the 1st term of the sequence we double 1 and then add 3
This gives us 2 x 1 = 2 then 2 + 3 = 5
To find the 2nd term of the sequence we double 2 and then add 3
This gives us 2 x 2 = 4 then 4 + 3 = 7
To find the 3rd term of the sequence we double 3 and then add 3
This gives us 2 x 3 = 6 then 6 + 3 = 9
To find the 4th term of the sequence we double 4 and then add 3
This gives us 2 x 4 = 8 then 8 + 3 = 11
To find the 5th term of the sequence we double 5 and then add 3
This gives us 2 x 5 = 10 then 10 + 3 = 13
So the first five terms of the sequence 'double and add 3' are:
5, 7, 9, 11, 13
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After making sequences using a rule in words we move on to being able to make a sequence using an algebraic rule
All sequences that increase in equal amounts are called linear sequences and these are the easiest types of sequences to make and to interpret algebraically.
The rule to most linear sequences is expressed with an 'n' as the unknown term. try to remember that n stands for number
On the next 2 pages is an example of how to create a sequence having been given the rule in an algebraic format.......
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Generating sequences using an algebraic rule e.g Calculate the first 5 terms of the sequence 5n + 3
To calculate the first number in the sequence we put n = 1 into the formula
To calculate the second number in the sequence we put n = 2 into the formula
To calculate the third number in the sequence we put n = 3 into the formula..... etc.... up to n = 5.
So if n = 1, 5n + 3 = (5 x 1) + 3
= 5 + 3
= 8
So if n = 2, 5n + 3 = (5 x 2) + 3
= 10 + 3
= 13
So if n = 3, 5n + 3 = (5 x 3) + 3
= 15 + 3
= 18
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So if n = 4, 5n + 3 = (5 x 4) + 3
= 20 + 3
= 23
So if n = 5, 5n + 3 = (5 x 5) + 3
= 25 + 3
= 28
This gives us the sequence 8, 13, 18, 23, 28
So if we wanted to know what the 75th term of the above sequence is we can put n = 75 into the formula....
if n = 75, 5n + 3 = (5 x 75) + 3
= 375 + 3
= 378
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Sometimes we are given the rule to a sequence in words and we have to work out previous terms.
This again involves working backwards doing the inverse (opposite) to the rule.
e.g John thinks of a number, he doubles it and then adds 7 and gets 49, what was his number?
Johns rule is 'double and add 7' Writing this out mathematically looks like: ? x2 +7 = 49
We work backwards through this and do the inverse operations each time.
? x2 +7 = 49
21 = ÷2 ­ 7 49
So as yu can see the answer is 21
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Sometimes we are given the sequence and have to figure out what the rule was algebraically.
All linear sequences have a rule which takes the form an ± b where a and b are numbers (a is the multiplier).
e.g find the rule of the following sequence in algebraic form:
8, 11, 14, 17, 20, 23 .........
To do this we need to consider the difference between the terms
This difference becomes the multiplier in the rule
We can see that the terms are increasing by 3 each time so 3 is our multiplier.
So we can now see that the rule so far is 3n ± b
Now we need to figure out the last bit of the rule.
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To do this we look at the first term and make it equal to the rule
this gives us: 3n ± b = 8
as 8 is the 1st term of the sequence this means that the n in the above statement is equal to 1.
This now gives us: (3x1) ± b = 8
3 ± b = 8
ask yourself: what number must I replace the b with to make this statement true?
The only thing that will work is 3 + 5 = 8
So: b = 5
This gives us the rule: 3n + 5
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Sometimes we are given a sequence where the second part of the rule is a subtraction.
e.g find the rule of the following sequence in algebraic form:
5, 12, 19, 26, 33, 40 .........
To do this we need to consider the difference between the terms
This difference becomes the multiplier in the rule
We can see that the terms are increasing by 7 each time so 7 is our multiplier.
So we can now see that the rule so far is 7n ± b
Now we need to figure out the last bit of the rule.
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To do this we look at the first term and make it equal to the rule
this gives us: 7n ± b = 5
As 5 is the 1st term of the sequence this means that the n in the above statement is equal to 1.
This now gives us: (7x1) ± b = 5
7 ± b = 5
ask yourself: what number must I replace the b with to make this statement true?
The only thing that will work is 7 ­ 2 = 8
So: b = ­2
This gives us the rule: 7n ­ 2
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A quadratic sequence is one where the highest power is 2.
e.g n2 is a quadratic sequence, as is 4n2 + 2n ­3 A quadratic sequence does not increase in equal amounts like a linear sequence.
It is possible to make a quadratic sequence (Level 7) by substituting values into the formula, just like linear sequences.
With a quadratic sequence you need to be careful and remember BODMAS (the order of calculations)
e.g make the first 5 terms of the sequence n2 + 3
when n = 1: n2 + 3 = (1 x 1) + 3 = 1 + 3 = 4
when n = 2: n2 + 3 = (2 x 2) + 3 = 4 + 3 = 7
when n = 3: n2 + 3 = (3 x 3) + 3 = 9 + 3 = 12
when n = 4: n2 + 3 = (4 x 4) + 3 = 16 + 3 = 19
when n = 5: n2 + 3 = (5 x 5) + 3 = 25 + 3 = 28
This gives the answer 4, 7, 12, 19, 28
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