Question #92422

2 Answers
Apr 1, 2017

Answer:

I got #33#

Explanation:

Call your numbers:
#n#
#n+1#
#n+2#
#n+3#
So we get:
#n+(n+1)+(n+2)+(n+3)=130#
rearrange and solve for #n#:
#4n+6=130#
#4n=124#
#n=124/4=31#
So the third in the sequence will be #33#

Apr 1, 2017

Answer:

#33# (see explanation)

Explanation:

Since the sum consists of #4# consecutive integers we can write the following and solve for #x#

#x+(x+1)+(x+2)+(x+3)= 130#

#4x+6=130 -> 4x+cancel(6-6)=130-6#

#4x=124 -> cancel(4/4)x=124/4#

#x=31#

Thus the four consecutive integers are as follows:

#31+(31+1)+(31+2)+(31+3)=130#

or

#31,32,33,34#

We can verify this by substituting #31# into the equation we used to solve for #x#

#31+32+33+34=130#

#130=130#

Finally, since we were asked to find third number of the sequence, we know that our third number is #33#