Question

Find 3 consecutive numbers such that 3 times the first is equal to 8 more than the sum of the other two.
a. 8, 9, 10
b. 9, 10, 11
c. 11, 12, 13
d. 12, 13, 14

Answer 1

Let the three numbers be x, x + 1, x + 2.

ATP,
3x = x + 1 + x + 2 + 8
3x = 2x + 11
x = 11

x = 11
x + 1 = 12
x + 2 = 13

Therefore, the required numbers are c.
11, 12, 13. Sol.

Thanks!