Question

Find the three consecutive numbers where the sum of twice of first number, thrice of second number and four times of third number is 182.

Answer 1

Let three consecutive numbers be x, x + 1 and x + 2.

Then according to question 2x + 3(x + 1) + 4 (x + 2) = 182

⇒ 2x + 3x + 3 + 4x + 8 – 182

⇒ 9x + 11 = 182

⇒ 9x = 182 – 11

⇒ 9x = 171

⇒ x = 171/9 = 19

Hence, the first number = 19 

second number = 19 + 1 = 20

and third number = 19 + 2 = 21