Find three consecutive integers such that, the sum of the first and four times the third is 98.
Thank you!
Numbers are:
A
B = A + 1
C = B + 1 = A + 1 + 1 = A + 2
The sum of the first and four times the third is 98 means;
A + 4 C = 98
A + 4 ∙ ( A + 2 ) = 98
A + 4 A + 8 = 98
5 A + 8 = 98
Subtract 8 to both sides
5 A = 90
A = 90 / 5 = 18
So numbers are:
A = 18
B = A + 1 = 18 + 1 = 19
C = A + 2 = 18 + 2 = 20
18 , 19 , 20