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

Hello Guest!

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

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

!

asinus
Sep 19, 2017