1. nth term
maths
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2. Introduction
Steps in discovering the nth term
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3. Write out the sequence in a table8 15 22 29
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4. Find the first difference
Difference is 7
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5. Is it constant
Yes, always 7
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6. Write down that n x difference
N x 7
or 7n
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7. Work out what you need to add or subtract to each term, after the multiplication, to make the sequence
1st
2nd
3rd
4th 8 15 22 29
7n + ____ = 8
When n is the first term multiply 7 x 1 + ___= 8
When n is the second term multiply 7 x 2 + ___= 15
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8. Then you should get the rule
When n is the first term multiply
7 x 1 + ___= 8
The rule is 7n + 1
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9. This rule 7n + 1 should work for all the terms
1st
2nd
3rd
4th 8 15 22 29
7n + 1 = 29 when n = 4
7 x 4 = = 29
So now I can use the rule to find the 100th term
7x = 701
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10. 1.Find the solution to the sequence
TRY THESE IN YOUR BOOKS
1.Find the solution to the sequence
15,19,23,27, difference is 4
21,25,29,33
13,22,31,40
19,24,29,34
23,25,27,29
6,13,20,27
2.What is the 20th term of each one
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11. If the sequence is negative ?
The same applies
Find the difference
1st
2nd
3rd
4th 8 5 2 -1
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12. 1st 2nd 3rd 4th 8 5 2 -1 Difference is -3 -3n + ____ = 8
When n is the first term multiply -3 x 1 + ___= 8
When n is the second term multiply -3 x 2 + ___= 5
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13. TRY THESE 1.Find the solution to the sequence 15,11,7,3 21,16,11,6
13,10,7,4
9,0,-9,-18
23,17,11,5
6,1,-4,-9
2.What is the 20th term of each one
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14. What if the first difference is not constant ?
2nd
3rd
4th
5th 3 6 11 18 27
The differences form a linear sequence
3,5,7,9
If this happens the rule will use n squared N2
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15. So the 2nd sequences goes up in 2’s
1st
2nd
3rd
4th 3 6 11 18 5
To get the rule for the nth term we need to square the term. So for the first try 1 x 1and then add/subtract and test.
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16. In this case it is simple
N2 + 2 gets to 3 if n = 1
N2 + 2 gets to 6 if n = 2
So N is the rule
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17. This pattern continues
1st
2nd
3rd
4th
5th 2 6 12 20 30
The first difference increases by 4
The second increases by 2
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18. You need to learn these rules
You'll need to learn the following rules: If the second difference is 2 you start with n2 If the second difference is 4 you start with 2n2 If the second difference is 6 you start with 3n2
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19. TRY THESE
2,5,10,17,26,37
4,7,12,19,28
3,9,17,27,39
2,8,18,32,50,72
8.10,14,20,28
75,81,89,99,111
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20. So must be n2 (thanks shu sam)
This one is tough !
3,9,17,27,39
1st sequence 6,8,10,12
2nd sequence 2,2,2,2
So must be n2 (thanks shu sam)
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21. Let’s just test that Term n = 2n2 + 3 find the first 4 terms
Term n = 3n2 + 5n find the first 4 terms
Term n = n2 - 2n find the first 4 terms
Term n = 4n2 + n find the first 4 terms
Term n = 3n2 + 4n find the first 4 terms
The n2 sequences goes up in 2’s the 2n2 sequence in 4’s the 3n2 sequence in 6’s
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22. We need the quadratic rule for a sequence
WHICH IS
Term n = an2 + bn + c
n = the term number
a = the second sequence divided by 2.
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23. Term n = an2 + bn + c Term 1 = a x 1 + b x 1 + c Term 1 = a + b + c
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24. Term n = an2 + bn + c 2a 3a + b 5a +b 7a + b Term 1st 2nd 3rd 4th
a + b + c
4a + 2b +c
9a + 3b + c
16a + 4b +c
1st Difference
3a + b
5a +b
7a + b
2nd Difference 2a
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25. 2a Term n = an2 + bn + c a = 1 and 3a + b = 6 so b must = 3
3rd
4th
terms
a + b + c
4a + 2b +c
9a + 3b + c
16a + 4b +c
1st Difference
3a + b
5a +b
7a + b
2nd Difference 2a
a = 1 and 3a + b = 6 so b must = 3
a + b + c = 3 so c = -1
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26. Term n = an2 + bn + c a = 1 and 3a + b = 6 so b must = 3
a + b + c = 3 so c = -1
Term n = 1 x x 1 - 1
Simplified n n - 1
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27. 3,9,17,27,39 1st sequence 6,8,10,12 2nd sequence 2,2,2,2
Term n = an2 + bn + c
A = 2 ÷ 2 = 1
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28. There is a formula nth term = a + (n - 1)d + 1⁄2(n - 1)(n - 2)c
This time there is a letter c which stands for the second difference (or the difference between the differences) and d is just the difference between the first two numbers.
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