Arithmetic means?

Question

Arithmetic means are inserted between 12 and 36 so that the sum of the first three means is to the sum of the last three means as 3:5. How many means are inserted?

Answer 1

7 Means are inserted. Here is how I calculated that:

Between each of these means is the same interval, which I'll call x.

12 + x, 12 + 2x, and 12 + 3x are the first 3 means, and their sum is 36 + 6x.

The last 3 are 36 - x, 36 - 2x, and 36 - 3x, and their sum is 108 - 6x.

Now we know these two are in the ratio 3:5, so that if we multiplied the first by 5 and the second by three, the two results equal each other. Thus,

5(36 + 6x) = 3(108 - 6x)

Or, 180 + 30x = 324 - 18x

Add 18x to both sides:

180 + 48x = 324

Subtract 180 from both sides:

48x = 144

Divide by 48:

x = 3

So the interval is 3. Between 12 and 36, inserting a mean every 3 numbers, there are 15, 18, 21, 24, 27, 30, and 33. That is 7.
7 means are inserted.

** Edit:
As for answer #1, the answerer makes the mistake of calling 12 the first mean and 36 the last mean. Your question clearly states that the means are *between* 12 and 36, and does not state that either 12 or 36 counts as a mean.

Answer 2

an=a1+(n-1)d

an-a1=(n-1)d
36-12=(n-1)d

24=(n-1)d

a1+a2+a3
--------------
an+an-1+an-2


=3/5

12+(12+d)+(12+2d)
----------------------
36+12+(n-2)d+12+(n-3)d

=3/5

Let the salt be in your soup.
simplify by yourself.
Then finding n is easy.