Question

Find five consecutive integers whose sum is 195

Answer 1

$n;\ n+1;\ n+2;\ n+3;\ n+4-five\ consecutive\ integers\\\\(n)+(n+1)+(n+2)+(n+3)+(n+4)=195\\n+n+1+n+2+n+3+n+4=195\\5n+10=195\ \ \ \ \ |subtract\ 10\ from\ both\ sides\\5n=185\ \ \ \ \ \ |divide\ both\ sides\ by\ 5\\n=37\\\\Answer:\boxed{37;\ 38;\ 39;\ 40;\ 41}$

Answer 2

To do this, come up with three numbers. These are n, n+1, n+2, n+3, and n+4.

To solve, you do this:

$n+n+1+n+2+n+3+n+4=195 \\ 5n+10=195 \\ 5n+(10-10)=(195-10) \\ 5n=185 \\ \frac{5n}{5} = \frac{185}{5} \\ n=37$

Then, substitute 37 into the numbers:

n=37
n+1=37+1=38
n+2=37+2=39
n+3=37+3=40
n+4=37+4=41

The five consecutive integers are 37, 38, 39, 40, and 41.