SOLUTION: If S1,S2,S3 be the sum of n,2n,3n terms respectively of an A.P.Then (a)S3=S1+S2 (b)S3=2(S1+S2) (c)S3=3(S2-S1) (d)S3=3(S1+S2) If Sn denotes the sum of first n terms of

Algebra ->  Sequences-and-series -> SOLUTION: If S1,S2,S3 be the sum of n,2n,3n terms respectively of an A.P.Then (a)S3=S1+S2 (b)S3=2(S1+S2) (c)S3=3(S2-S1) (d)S3=3(S1+S2) If Sn denotes the sum of first n terms of       Log On


   



Question 981006: If S1,S2,S3 be the sum of n,2n,3n terms respectively of an A.P.Then (a)S3=S1+S2

(b)S3=2(S1+S2) (c)S3=3(S2-S1) (d)S3=3(S1+S2)


If Sn denotes the sum of first n terms of an A.P and S2n=3Sn,then S3n/Sn=




Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If S%5B1%5D , S%5B2%5D , and S%5B3%5D are the sums of n , 2n , 3n terms respectively of an A.P., then highlight%28S%5B3%5D=3%28S%5B2%5D-S%5B1%5D%29%29 .

ONE WAY TO SOLVE THE PROBLEM is to applying somewhat involved algebraic manipulations,
and (possibly memorized) formulas derived from (possibly forgotten, possibly never fully understood) concepts:
For an A.P. with first term a%5B1%5D and common difference d ,
Sum of first m terms =2a%5B1%5D%2B%28m-1%29d%29m%2F2 <---> S%5Bm%5D=%282ma%5B1%5D%2Bm%28m-1%29d%29%2F2
For m=n , we have S%5B1%5D=%282na%5B1%5D%2Bn%28n-1%29d%29%2F2=%282na%5B1%5D%2B%28n%5E2-n%29d%29%2F2 .
For m=2n , we have S%5B2%5D=%284na%5B1%5D%2B2n%282n-1%29d%29%2F2=%284na%5B1%5D%2B%284n%5E2-2n%29d%29%2F2 .
For m=3n , we have S%5B3%5D=%286na%5B1%5D%2B3n%283n-1%29d%29%2F2=%286na%5B1%5D%2B%289n%5E2-3n%29d%29%2F2 .
,
so the three choices with S%5B1%5D%2BS%5B2%5D do not work.
On the other hand,
,
so

ANOTHER WAY TO LOOK AT THE PROBLEM
involves visualizing what those sums are made of and how they are related,
relying more on concepts than on formulas,
and using easier arithmetic rather than advanced algebra.
Unfortunately, describing a clear reasoning based on basic concepts often takes a lot of words (or a lot of algebraic expressions).
Let the terms of the A.P. be a%5B1%5D , a%5B2%5D=a%5B1%5D%2Bd , a%5B3%5D=a%5B1%5D%2B2d , etc.
S%5B1%5D=a%5B1%5D%2Ba%5B2%5D%2B%22...%22%2Ba%5Bn%5D
S%5B2%5D=S%5B1%5D%2Ba%5Bn%2B1%5D%2Ba%5Bn%2B2%5D%2B%22...%22%2Ba%5B2n%5D
S%5B3%5D=S%5B2%5D%2Ba%5B2n%2B1%5D%2Ba%5B2n%2B2%5D%2B%22...%22%2Ba%5B3n%5D
We know that each term is the one before plus d ,
and the difference between two terms is d times the difference between the term numbers,
as in a%5Bk%5D=a%5B1%5D%2B%28k-1%29d <---> a%5Bk%5D-a%5B1%5D%2B%28k-1%29d ,
so a%5Bn%2B1%5D=a%5B1%5D%2Bnd ,
but that applies more generally, not just when one of those two terms is a%5B1%5D ,
so a%5Bn%2B2%5D=a%5B2%5D%2Bnd , a%5Bn%2B3%5D=a%5B3%5D%2Bnd , and so on up to
a%5B2n%5D=a%5Bn%2Bn%5D=a%5Bn%5D%2Bnd .
So, .
We can write S%5B2%5D as a function of X=S%5B1%5D and Y=n%5E2d :
S%5B2%5D=2S%5B1%5D%2Bn%5E2d <--> S%5B2%5D=2X%2BY .
By a similar reasoning,
since a%5B2n%2B1%5D=a%5B1%5D%2B2nd , a%5B2n%2B2%5D=a%5B2%5D%2B2nd , and so on, up to a%5B3n%5D=a%5B2n%2Bn%5D=a%5Bn%5D%2B2nd ,
,
so S%5B3%5D=S%5B2%5D%2BS%5B1%5D%2B2n%5E2d ,
and system%28S%5B2%5D=2S%5B1%5D%2Bn%5E2d%2CS%5B3%5D=S%5B2%5D%2BS%5B1%5D%2B2n%5E2d%29 ---> S%5B3%5D=2S%5B1%5D%2Bn%5E2d%2BS%5B1%5D%2B2n%5E2d=3S%5B1%5D%2B3n%5E2d .
We can write S%5B3%5D as a function of X=S%5B1%5D and Y=n%5E2d :
S%5B3%5D=3S%5B1%5D%2B3n%5E2d <--> S%5B3%5D=3X%2B3Y .
Looking at system%28S%5B1%5D=X%2CS%5B2%5D=2X%2BY%2CS%5B3%5D=3X%2B3Y%29 ,
it is clear that none of the formulas with S%5B1%5D%2BS%5B2%5D=3X%2BY will work,
but S%5B2%5D-S%5B1%5D=X%2BY--->3%28S%5B2%5D-S%5B1%5D%29=3%28X%2BY%29=3X%2B3Y=S%5B3%5D

NOW FOR THE EASY PROBLEM:
If Sn denotes the sum of first n terms of an A.P.,
and S%5B2n%5D=3S%5Bn%5D , then S%5B3n%5D%2FS%5Bn%5D= ?
We proved above that if S%5B1%5D , S%5B2%5D , and S%5B3%5D are the sums of n , 2n , 3n terms respectively of an A.P., then highlight%28S%5B3%5D=3%28S%5B2%5D-S%5B1%5D%29%29 .
Using the easier symbols from the problem above,
the new question is
if S%5B2%5D=3S%5B1%5D , then S%5B3%5D%2FS%5B1%5D= ?
S%5B3%5D=3%28S%5B2%5D-S%5B1%5D%29=3S%5B2%5D-3S%5B1%5D , so
system%28S%5B3%5D=3S%5B2%5D-3S%5B1%5D%2CS%5B2%5D=3S%5B1%5D%29--->S%5B3%5D=3%283S%5B1%5D%29-3S%5B1%5D=9S%5B1%5D-3S%5B1%5D=6S%5B1%5D--->S%5B3%5D%2FS%5B1%5D=6S%5B1%5D%2FS%5B1%5D=highlight%286%29