Find three consecutive numbers such that triple the last number gives the same result as twice the sum of the first two numbers

Answer 1

Consecutive numbers are like ( 1, 2, 3..) or (18, 19, 20) or (111, 112, 113..)

1st consecutive # is = x

2nd consecutive # is = x + 1 (because the next number in the consecutive series increases by 1 from the first #)

3rd consecutive # is = x + 2 (because the next number in the consecutive series increases by 2 from the first #)

Now we apply the question as an equation:

Triple the last number = 3 (x + 2)

Twice the sum of the first 2 numbers is = 2 (x + x + 1)

Now "triple the last number gives the same result as twice the sum of the first two numbers"

Means =

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

2(2x + 1) = 3x + 6

4x + 2 = 3x + 6

Combine like terms

x = 4 (first number)

x + 1 = 4 + 1 = 5 (second number)

x + 2 = 4 + 2 = 6 (third number)

So, the consecutive numbers are 4, 5, 6

Check:

2 (4 + 4 + 1) = 3 (4 + 2)

2(9) = 3(6)

18 = 18