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Expert: - 12/24/2011


Question
The first term of an arithmetic sequence is 3p+5, where p is a positive integer. The last term is 17p+17, and the common difference is 2.

Show that the sum of the sequence is divisible by 14 only when p is odd.

Could you please help me work through this question? I am unsure what to do. Thanks!

Answer
The terms go from 3p+5 to 17p+17, and they are by 2, so take the last term minus the first term, divide by 2, and add 1 to get the number of terms.  That is, 17p+17 - 3p+5 = 14p+12,
and that over 2 is 7p+6, and plus 1 is 7p+7 = 7(p+1).

The average is the just the first plus the last, all over 2, which is (3p+5 + 17p+17)/2,
and that is 10p+11.

This sum of terms be the number times the average, which is 7(p+1)(10p+11).

This number is divisible by 7 for any p, but to be divisible by 14, we need one of the other two terms divisible by 2.  Note that 10p+11 is always odd, so it is never divisible by 2.
For the p+1,  it is divisible by 2 any time p is odd, which makes the result divisible by 2 and 7.

Thus, for any odd n, the sum is divisible by 14.
For n being even, the sum is an odd multiple of 7.  

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