If a is the sum of x consecutive positive integers. b is the sum of y

The sum n consecutive integers is give by: \(Sum=\frac{(2a_1+n-1)*n}{2}\) (check Number Theory chapter of Math Book for more: math-number-theory-88376.html);

Note that:
when \(n=even=2*odd\), so when \(n\) (# of consecutive integers) is even but not a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*odd)}{2}=odd*odd=odd\);
when \(n=even=2*even\), so when \(n\) is a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*even)}{2}=odd*even=even\);

That's because a set of even number of consecutive integers has half even and half odd terms. The sum of even terms is obviously even. As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);

So, the sum of 10 (not a multiple of 4) consecutive integers will be odd (the sum of 5 even and 5 odd integers) and the sum of 4 (multiple of 4) consecutive integers will be even (the sum of 2 even and 2 odd integers), so option D is not possible.

Answer: D.

Hope it's clear.