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

Algebra ->  Sequences-and-series -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
The n arithmetic means, a%5B1%5D , a%5B2%5D , a%5B1%5D , ...,a%5Bn%5D ,
are equally spaced by a common difference d , forming an arithmetic sequence (or arithmetic progression):
a%5B0%5D=75 ,
a%5B1%5D=75-d ,
a%5B2%5D=75-2d ,
a%5B3%5D=75-3d ,
a%5B4%5D=75-4d ,
................... ,
a%5Bn%5D=75-nd , and
19=75-%28n%2B1%29d .
We do not know n , and we do not know d ,
but we have one equation relating them:
19=75-%28n%2B1%29d <---> %28n%2B1%29d=75-19 <---> %28n%2B1%29d=56 .
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
a%5Bn%5D=19%2Bd ,
a%5Bn-1%5D=19%2B2d ,
a%5Bn-2%5D=19%2B3d ,
a%5Bn-3%5D=19%2B4d ,
a%5Bn-4%5D=19%2B5d ,
and so on.
So, a%5Bn-3%5D%2Ba%5Bn-4%5D=%2819%2B4d%29%2B%2819%2B5d%29=38%2B9d , and
we knew that a%5B3%5D=75-3d and a%5B4%5D=75-4d , so
a%5B3%5D%2Ba%5B4%5D=%2875-3d%29%2B%2875-4d%29=150-7d .
The ratio of those sums is
%28150-7d%29%2F%2838%2B9d%29=61%2F37
Solving that equation for d we get
%28150-7d%29%2F%2838%2B9d%29=61%2F37
37%28150-7d%29=61%2838%2B9d%29
37-150-37%2A7d=61%2A38%2B61%2A9d
5550-259d=2318%2B549d
5550-2318=259d%2B549d
3232=808d
3232%2F808=d ---> d=4
Now, we plug that value for d into the equation
%28n%2B1%29d=56 that we had, and get
%28n%2B1%29%2A4=56 <--> n%2B1=56%2F4 <--> n%2B1=14 <--> n=14-1 <--> highlight%28n=13%29