SOLUTION: if "n" arithmetic means�were�inserted between 75 and 19 such that the sum of the third and the fourth means to the sum of the means whose order are (n-3)�and (n-4) equals 61: 37 ev

Question 957473: if "n" arithmetic means�were�inserted between 75 and 19 such that the sum of the third and the fourth means to the sum of the means whose order are (n-3)�and (n-4) equals 61: 37 evaluate n. .... I know that n equals 13 but i don't know how
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
The arithmetic means, , , , ..., ,
are equally spaced by a common difference , forming an arithmetic sequence (or arithmetic progression):
,
,
,
,
,
................... ,
, and
.
We do not know , and we do not know ,
but we have one equation relating them:
<---> <---> .
We just need another equation.
From the given ratio we need to get the other equation.
The easiest way I could think is to look at that arithmetic sequence/progression in reverse.
Starting from the other end
,
,
,
,
,
and so on.
So, , and
we knew that and , so
.
The ratio of those sums is

Solving that equation for we get

--->
Now, we plug that value for into the equation
that we had, and get
<--> <--> <--> <-->
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