Find the mean and the median in the following arithmetic sequences:

Question

A: 1, 3, 5, 7, 9
B: 1, 3, 5, 7, 9, . . . ., 199
C: 7, 10, 13, 16, . . ., 607

Answer 1

OK: I'll show you how to find n (how many numbers in the sequence) as well. You'll most likely need to now how.

A. (1+3+5+7+9)/5 is a mean of 5 and a median of 5.
- This one is simple enough, just add them and divide by how many there are.
And of course, look in the middle for the median.

B. (1+3+5+7+9...199) is a mean of
-A little more complicated, the series has a common difference: d of 2 and goes from 1 to 199. We don't know how many numbers are in this series though.
To solve for this:n, use the formula:

a1+d(n-1)=ax
a1+2(n-1)=ax
1+2(n-1)=199
1+2n-2=199
2n-2=198
2n=200
n=100

We now know that there are 100 numbers in the series.

Solving for mean and median:

(a1+ax)/2, simply add the first and last and divide by 2.

(1+199)/2 = 200/2 = 100 is the mean.

The median of an arithmetic sequence is just the average, so it is 100 as well.

C. (7,10,13,16...607)
-Same idea as B.

The common difference: d is 3.

Finding n....

a1+d(n-1)=ax
a1+3(n-1)=ax
7+3(n-1)=607
7+3n-3=607
3n-3=600
3n=603
n=201

Now to find the mean and median, which are the same in an arithmetic sequence:

(a1+ax)/2 = (7+607)/2 = 614/2 =307 is the mean and median.

SO:
A. Mean and Median: 5
B. Mean and Median: 100
C. Mean and Median: 307

Adding the first and last and dividing by 2 works because each number is an equal distance apart.