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Q: What is the nth term for the sequence 18 10 2 6 14?

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The nth term would be -2n+14 nth terms: 1 2 3 4 Sequence:12 10 8 6 This sequence has a difference of -2 Therefore it would become -2n. Replace n with 1 and you would get -2. To get to the first term you have to add 14. Therefore the sequence becomes -2n+14. To check your answer replace n with 2, 3 or 4. You will still obtain the number in the sequence that corresponds to the nth term. :)

The nth term in this sequence is 4n - 2.

They are: nth term = 6n-4 and the 14th term is 80

It is: nth term = -4n+14

Clearly here the nth term isn't n25.

The nth term is: 3n+2 and so the next number will be 20

t(n) = 10 - 6n where n = 1, 2, 3, ...

This is an arithmetic sequence which starts at 14, a = 14, and with a common difference of -1, d = -1. We can use the nth term formula an = a + (n - 1)d to get an = 14 + (n - 1)(-1) = 14 - n + 1 = 15 - n.

25

You can see that all the numbers go up by 7. This means that the first part of the nth term rule for this sequence is 7n. Now, you have to find out how to get from 7 to 3, 14 to 10, 21 to 17 ... this is because we are going up in the 7 times table. To get from the seventh times table to the sequence, you take away four. So the answer is : 7n-4

If the nth term is n*7 then the first 5 terms are 7, 14, 21, 28, 35.

It is: 26-8n

It is: nth term = 6n-4

t(n) = n2 + n + 8

1,7,13,19

The nth term of this sequence is 3n + 4

5, 8, 11, 14 and 17.

f(n) = 14 - 6n -6n+20

t(n) = 6 + 8n

14+9n

First look for the difference between the terms, for example the sequence: 5, 8, 11, 14... has a difference of 3. This means the sequence follows the 3 times table - i.e. 3n Now since we need the first term to be 5 we add 2 to our rule to make it work. So the nth term of this sequence is 3n + 2.

42

-34 would be the 15th term.

By varying the parameters of a quartic polynomial, the nth term can be made whatever you like. But, taking the simplest solution, Un = 2 - 4n for

The simplest rule is Un = 26 - 8n for n = 1, 2, 3, ... but it is always possible to fit a polynomial of degree 5 to the given sequence of numbers along with any sixth number.