The sum of four consequtive positive integers is "X". In terms of "X", what is the sum of the second and third integers?

a) x/2 b) x-12/4 c) x-6/2 d) 2x+6

*a*,

*a*+ 1,

*a*+ 2,

*a*+ 3.

So the sum of four consecutive positive integers is

*X*, so we have;

$a+a+1+a+2+a+3=X\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+3+2a+3=X\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(2a+3\right)=X\phantom{\rule{0ex}{0ex}}\Rightarrow \left(2a+3\right)=\frac{X}{2}$

And the sum of second and third consecutive integers =

*a*+ 1 +

*a*+ 2 = 2

*a*+ 3

Therefore, the sum of second and third consecutive integers = 2

*a*+ 3 = $\frac{X}{2}$.

So option (a) is correct.

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