SOLUTION: 1,7,13.....are in ap.can the sum of any terms be 175?justify

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Question 994585: 1,7,13.....are in ap.can the sum of any terms be 175?justify
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
For an AP (arithmetic progression or arithmetic sequence) with first term a%5B1%5D and common difference d ,
the sum of the first n terms is
%28a%5B1%5D%2Ba%5Bn%5D%29%2An%2F2 or
.

The sum of the first n terms cannot be 175.

ONE APPROACH:
For the AP 1, 7, 13, ... , with system%28a%5B1%5D=1%2C+d=7-1=6%29 ,
the sum of the first n terms is

n%283n-2%29=175-->3n%5E2-2n=175-->3n%5E2-2n-175=0
The solution to that equation is not a valid n , because it is not an integer:
.

ANOTHER APPROACH:
The sum of the first n terms would be %281%2Ba%5Bn%5D%29%2An%2F2=%28%281%2Ba%5Bn%5D%29%2F2%29%2An
That means that %281%2Ba%5Bn%5D%29%2F2 and n are integers that multiply to yield 175 .
We know that %281%2Ba%5Bn%5D%29%2F2%3En ,
because %28a%5B1%5D%2Ba%5B2%5D%29%2F2=%281%2B7%29%2F2=8%2F2=4%3E2 ,
and for each unit increase in n , %281%2Ba%5Bn%5D%29%2F2 increases by d%2F2=6%2F2=3 .
That makes the possible combinations of factors of 175 are
5%2A35=175 , which would mean system%28n=5%2C1%2Ba%5Bn%5D%29%2F2=35%29 , and
7%2A25=175 , which would mean system%28n=7%2C1%2Ba%5Bn%5D%29%2F2=25%29 .
Neither option works:
n=5 means a%5Bn%5D=a%5B5%5D=1%2B4%2A6=1%2B24=25 and %281%2Ba%5Bn%5D%29%2F2=%281%2Ba%5B5%5D%29%2F2=%281%2B25%29%2F2=13 .
n=7 means a%5Bn%5D=a%5B7%5D=1%2B6%2A6=1%2B36=37 and %281%2Ba%5Bn%5D%29%2F2=%281%2Ba%5B7%5D%29%2F2=%281%2B37%29%2F2=19 .

A DIFFERENT INTERPRETATION OF THE PROBLEM:
The sum of n consecutive terms (not starting from a%5B1%5D=1 ) could be 175 :
7%2B13%2B19%2B25%2B31%2B37%2B43=175 .