Question

Four consecutive odd intergers can be represented by n, n+2, n+4, and n+6. If the sum of 4 consecutive odd integers is 56, what are the integers?

Answer 1

first we have to arrange them as a equation
- n+(n+2)+(n+4)+(n+6)
=4n +12=56
=4n=56-12
=4n=44
=n=44/4
n=11
so the integers are = n,n+2,n+4,n+6
=11,11+2,11+4,11+6
=11,13,15,17
but because it is odd no.
it will be =n+1)+(n+2+1)+(n+4+1)+(n=6=1)
=4n+16
=41n=40
=n=10
=11,13,15,17

Answer 2

the numbers are.,
n+n+2+n+4+n+6=56
4n+12=56
4n=44
n=11
put n=11 in the eqn,
n=11
n+2=11+2=13
n+4=11+4=15
n+6=11+6=17
the numbers are 11,13,15,17