math

If the sum of four consecutive numbers is n, what is the first one?

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The first one is x.

$x+(x+1)+(x+2)+(x+(4-1))=n\\ n=4x+1+2+3\\ x=\frac{n-(1+2+3)}{4}$

$x = \frac{n-6}{4}$

If the sum of z consecutive numbers is n, what is the first one?

$x+(x+1)+(x+2)\ +\ ...\ +\ (x+z-1)=n$

$\frac{(x+(x+z-1)}{2}\times z=n\\ \frac{2x+z-1}{2}\times z=n\\ 2x+z-1=\frac{2n}{z}\\ 2x=\frac{2n}{z}-z+1\\ \color{blue}x=(\frac{2n}{z}-z+1)/2\\ \color{blue}general\ formula$   

sample

4+5+6+7+8+9   z=6   n=39    x=4

$x=(\frac{2n}{z}-z+1)/2=(\frac{2\cdot39}{6}-6+1)/2=4$.

  !