The cool thing about knowing that a set of numbers is an arithmetic sequence, is that, once I know the number that I am constantly adding (or subtracting), I can find any term in the sequence. I can even find the sum of all the terms in my sequence.
Let's look at a sequence we had in class today
(Remember, adding all these numbers gave us the number of diagonals in a 20 sided figure) :
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
Finding the nth Term
For starters, it looks like we're adding 1 to each previous term. Another way to say this is to say that the common difference (d) is 1.
And this can be broken down as follows:
1st Term = a1 = 2
2nd Term = a1 + d = a2 = 2 + 1 = 3
3rd Term = a2 + d = a3 = 3 + 1 = 4
4rd Term = a3 + d = a4 = 4 + 1 = 5
but rember, a2 can be written as: a1 + d
so the 3rd Term can actually be written as = a2 + d = a1 + d + d = a1 + 2d =2 + 1 + 1 = 4
and the 4rd Term can be written as = a3 + d = a1 + d + d + d = a1 + 3d =2 + 1 + 1 + 1 = 5
....so, the bottom, line is, you always only need the first term, and the common difference to find any number in a series. However, take care to notice something else... in the fourth term, we added the difference 3 times, and the third term we added 2 times...it's always one less than the current term we are on
The bottom line
So, to find the nth term, take the first term, and add the difference multiplied by 1 less than the number of terms (another way to look at it is we already have one term, we need to add the same number to get all the rest):
nth term = a1 + (n-1)d
Finding the sum of the first n Terms
The sum of all n terms in an arithmetic sequence is called an arithmetic series.
Let's look at a slightly simpler problem to understand this:
Look at these numbers: 1, 2, 3, 4, 5
Notice that the middle number is 3
Also notice that the average of the first and last terms and well as the two next inner terms, I would also get three:
(1+5)/2 = 3
(2+4)/2 = 3
This is because the numbers are evenly spaced, so the mean and the median (middle number) are always the same.
Notice also, that if I took that mean number and added it 5 times, I get the same answer as if I added all the terms in the original series
3 + 3 + 3 + 3 + 3 = 15 = 1 + 2 + 3 + 4 + 5
OR 3 * 5 = 15
The bottom line
Therefore, if we have the first and last terms, and we find the average, we can just multiply by the number of terms to sum it up.
n * [(a1 + a2)/2]
So...back to the question from today.
In short, how many diagonals does a 20 sided polygon have?
If we started just knowing the number 20 for the sides and the first term, we could reason that every shape after 3 has diagonals... so, we are looking to find the 17th (ie...20-3) term.
That would be
a1 + (n-1)d
= a1 + ((20-3)-1)d
= 2 + 16 (1)
= 18, the 17th term is 18
And if we wanted to sum up all those terms:
n * [(a1 + a2)/2] ...but first some fancy footwork...
While n = 20, we need to add only 17 terms, so let's represent that as (n-3).... 20 - 3 = 17
Also, the last term is 2 less than our number of sides, so let's call that (n - 2) .... 20 - 2 = 18 (last term)
And the first term is two, but we can rewite that using the last term
[n - (n-2)]....20 -(20-2) = 20 -18 =2 (first term)
...So we found the average of (n-2) and [n - (n-2)]
(n - 2 + n -n + 2)/2 = (n - n + n - 2 + 2)/2 = n /2
So our formula looks like
(n-3)*n/2
= (20-3) * 20/2
= 17 * 10
= 170
Just to check, notice that if we added all the numbers in the series we started with:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18...We would get 170.
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CHALLENGE PROBLEM:
Look back at the picture at the top of this post.
Context: A queen charges each person in her kingdom three gold coins a day. Every day, some new (unlucky wanderer) enters her kingdom and so her collection of coins is always three more than the day before.
1) How many coins does she collect on the 100th day
2) How many coins, in all, has she collected from day 1 to day 100?
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