SOLUTION: If the sum of n terms of two arithmetic progressions is in the ratio (3n+8): (7n+15) find the ratio of their 12th term.

Algebra ->  Sequences-and-series -> SOLUTION: If the sum of n terms of two arithmetic progressions is in the ratio (3n+8): (7n+15) find the ratio of their 12th term.      Log On


   

Question 1162673: If the sum of n terms of two arithmetic progressions is in the ratio (3n+8): (7n+15) find the ratio of their 12th term.
Found 3 solutions by solver91311, greenestamps, ikleyn:
Answer by solver91311(24713) About Me  (Show Source):
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I see your post got an empty response from another tutor. That usually means that tutor started on a response and decided to quit.

You will need to re-post.

This sounds as if it could be an interesting problem. However, the statement of the problem is grammatically incorrect; so we don't know what the problem really is.

The information as given in your post says

"... the sum of ...... is in the ratio (3n+8):(7n+15)..."

That is mathematical nonsense: a sum is a single number, not a ratio.

I suspect the ratio is to be between the sums of n terms of the two arithmetic progressions; but I'm not going to spend a lot of time working on a problem if I'm not sure what the question is.

Please re-post stating the problem correctly.

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The problem's formulation is crystally clear.

            See my solution below. 


Let a is the first term and d is the common difference of the first AP.

Let b is the first term and e is the common difference of the second AP.


Then


    ( (n/2)*(2a+(n−1)*d) ) / ( (n/2)*(2b+(n−1)*e) ) = (3n+8)/(7n+15)

or


    (2a+(n−1)*d)/(2b+(n−1)*e) = (3n+8)/(7n+15)

or

    ((2a-d) + nd)/((2b-e)+ed) = (3n+8)/(7n+15).



It means that up to a coefficient of proportionality


    2a−d = 8  and  d = 3,  which implies  2a = 11, d =3,  or  a = 5.5, d = 3


also  

    2b−e = 15  and  e = 7,  which implies  2b = 22, e = 7, or  b = 11, e = 7.



Now  the 12-th terms of the progressions are (up to proportionality)

    5.5 + 11*3 = 38.5  and  11 + 7*11 = 88


and their ratio is

    38.5%2F88 = 77%2F176 = 7%2F16.    ANSWER

Solved.