Expert: Ahmed Salami - 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
Hi James,
If the first term of an arithmetic sequence is 3p+5 and its common difference is 2, then the nth term (or last term) is given by;
T(n) = (3p+5) + (n-1).2
= 3p + 5 + 2n - 2
= 3p + 2n + 3
But in this case the last term is 17p + 17, so
17p + 17 = 3p + 2n + 3
14p + 14 = 2n
n = 7p + 7
Now, the sum of a sequence to the nth term is given by;
S(n) = (n/2)(a + l)
where a and l are the first and last terms respectively.
So,
S(n) = (n/2)(3p + 5 + 17p + 17)
= (n/2)(20p + 22)
= n(10p + 11)
= (7p + 7)(10p + 11)
and we can see that the term 7p+7 would only be divisible by 14 if p were odd.

To clearly show it, if p were even then we can write it as p = 2k and
7p + 7 = 7(2k) + 7
= 14k + 7
which wont be divisible by 14 since only 14k is and not 7.

And just to be certain, if p were odd then we can write it as p = 2k - 1 and
7p + 7 = 7(2k - 1) + 7
= 14k
which is clearly divisible by 14 and hence S(n) also is.

Regards