# Is there an nth term for the sequence 1, 5, 10, 15, 20, 25?

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I can't figure it out. It seems like an arithmetic sequence, but the difference from 1 to 5 isn't the same as the rest.

I can't figure it out. It seems like an arithmetic sequence, but the difference from 1 to 5 isn't the same as the rest.

##### 1 Answer

A possible formula is:

#a_n = 5(n-1) + 0^(n-1)#

Another is:

#a_n = 1/120 (-n^5 + 20 n^4 - 155 n^3 + 580 n^2 - 444 n + 120)#

#### Explanation:

You are correct in asserting that this looks like an arithmetic sequence apart from the first element.

The differences between successive elements are:

#4, 5, 5, 5, 5#

So there is no common difference.

This sequence does not have a common ratio either, the ratios being:

#5, 2, 3/2, 4/3, 5/4#

So it is neither an arithmetic sequence nor a geometric one.

In the absence of any other information about the given sequence, we cannot say how it would continue.

If the remainder of the sequence is arithmetic, then we could write a formula for the

#a_n = 5(n-1) + 0^(n-1)#

...assuming the convention that

**Polynomial matching**

Alternatively, we can match the given terms with a quintic polynomial as follows:

Write down the original sequence:

#color(blue)(1), 5, 10, 15, 20, 25#

Write down the sequence of differences between consecutive terms:

#color(blue)(4), 5, 5, 5, 5#

Write down the sequence of differences between those terms:

#color(blue)(1), 0, 0, 0#

Write down the sequence of differences between those terms:

#color(blue)(-1), 0, 0#

Write down the sequence of differences between those terms:

#color(blue)(1), 0#

Write down the sequence of differences between those terms:

#color(blue)(-1)#

Having arrived at a constant sequence (of one term), we can use the initial term of each of these sequences as coefficients to give us a formula:

#a_n = color(blue)(1)/(0!)+color(blue)(4)/(1!)(n-1)+color(blue)(1)/(2!)(n-1)(n-2)+color(blue)(-1)/(3!)(n-1)(n-2)(n-3)+color(blue)(1)/(4!)(n-1)(n-2)(n-3)(n-4)+color(blue)(-1)/(5!)(n-1)(n-2)(n-3)(n-4)(n-5)#

#color(white)(a_n) = 1/120 (-n^5 + 20 n^4 - 155 n^3 + 580 n^2 - 444 n + 120)#