Let,
1st integer = \(m\)
2nd integer = \(m+1\) and so on
5th integer = \(m+4\)
6th integer = \(m+5\)

average of the consecutive five integers = \(\frac{m+m+1+m+2+m+3+m+4}{5} = \frac{5(m+2)}{5} = m+2\)
average of the consecutive six integers = \(\frac{5m +10 +m +5}{6} = \frac{6m + 15}{6} = m +2.5\)
therefore the difference = \(m+2 - m - 2.5 = -0.5\)
option B
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Let m = 0

Then, we have set 1 {0,1,2,3,4) and set 2 (0,1,2,3,4,5)
Now, mean1 - mean2 = (0+1+2+3+4)/5 - (0+1+2+3+4+5)/6 = 2 - 15/6 = 2 - 2.5 = -1/2

--> B