Question

three consecutive integers are such that when they are taking in increasing order and multiplied bye 2 and 3 and 4 respectively they add up to 74 find this number

Answer 1

Let the integers be a, a + 1 and a + 2

According to question,

2 (a) + 3(a +1) + 4(a+ 2) = 74

=> 2a + 3a + 3 + 4a + 8 = 74

=> 9a + 11 = 74

=> 9a = 63

=> a = 7

First integer = 7

Second integer = 7 +1 = 8

Third integer = 7 + 2 = 9

Answer 2

let the integers be x, x+1, x+2
given
2x + 3(x+1) + 4(x+2) = 74
= 2x +3x +4x +3+8 = 74
= 9x = 74- 11
= 9x = 63
= x = 63/9 = 7

so the numbers are 7,8 and 9