What is the greatest positive integer that must divide the sum of the first ten terms of any arithmetic sequence whose terms are positive integers?
I believe that the answer to this question is 5.
This is because in any arithmetic sequence, there are 5 pairs of equal sums. So if the sum is any natural number(positive integer), then the 5 pairs' sum is also divisible by 5.
Hope this makes sense.
The terms of an AP are
a is the first term and d is what keeps getting added to give the next term
if a = -3 and d=5 the sequence would be -3, 2, 7, 12 etc see it starts at -3 and 5 keeps getting added.
With yours a is a positive integer and to make it easy d can be too
so the first 10 the sequence is
a, a+d, a+2d, .......a+9d
Add all those terms up and you get 10a+ 1d+2d+3d+4d+5d+6d+7d+8d+9d = 10a+ 45
What can be factorized out of that?
I am with my new account now
Thank you Melody! I think I got it! I greatly appreciate all of you who helped me with this problem!