Question #fcc37

2 Answers
Dec 7, 2017

Answer:

#53#

Explanation:

let#" "n" "#be the middle number

then we have
1st number#=n-2#

2nd number#=n-1#

4th number #n+1#

5th number#=n+2#

#(n-2)+(n-1)+n+(n+1)+(n+2)=270#

#n-cancel(2)+n-cancel(1)+n+n+cancel(1)+n+cancel(2)=270#

#5n=270#

#n=270/5=54#

#2nd" number "=54-1=53#

Answer:

Entire sequence: #52, 53, 54, 55, 56#
Second number in the sequence: #53#

Explanation:

If the sum of consecutive integers is equal to #270#, we can write the first integer as #x#, the 2nd as #x+1# (because they are consecutive), etc.

  • First integer: #x#
  • Second integer: #x+1#
  • Third integer: #x+2#
  • Fourth integer: #x+3#
  • Fifth integer: #x+4#

Notice with each integer, you are increasing the number by #1#. If that seems confusing, think about it this way:

If the first integer was #x#, the second will be #x+1#. With the third integer, we still increase by one from the previous integer:

#(x+1) ul(+1) = x+2#

Let's set up our equation:

#x+(x+1)+(x+2)+(x+3)+(x+4)= 270#

We have a good deal of terms to combine, so let's do that.

#x+x+x+x+x=5x#

#1+2+3+4= 10#

From this, we can make our new equation

#5x+10=270#

Subtract #10# from both sides to get

#5x=260#

Divide both sides by #5# to get

#x=52#

This isn't our answer, however. We found what #x# is (the first integer), but we want to know the second number in this sequence is, and to do that, just plug #52# into our expression for the second integer, #(x+1)#

#52+1= 53#

The second integer is #53#.