Answer:
38,39 and 40
Step-by-step explanation:
Given: the sum of 3 consecutive integral numbers is 117
To Find: Find the numbers.
Solution:
Let the three consecutive numbers be = [tex]\text{x}[/tex]
[tex]\text{x}+1[/tex]
[tex]\text{x}+2[/tex]
As given in question,
the sum of these three consecutive integral number is = [tex]117[/tex]
now,
[tex]\text{x}+\text{x}+1+\text{x}+2=117[/tex]
[tex]3\text{x}+3=117[/tex]
[tex]3\text{x}=114[/tex]
[tex]\text{x}=\frac{114}{3}[/tex]
[tex]\text{x}=38[/tex]
first number is 38, other two numbers are
[tex]\text{x}+1=39[/tex]
[tex]\text{x}+2=40[/tex]
To verify,
[tex]38+39+40=117[/tex]
Therefore three consecutive integrals whose sum is 117 are [tex]38,39[/tex] and [tex]40[/tex]
