Question

Find three consecutive integers such that the sum of the first, twice the second, and three times the third is 76.​

Answer 1

n + 2(n+1) + 3(n+2) = -76

Answer 2

Step-by-step explanation:

let the consecutive numbers be x, x+1,x+2 respectively.

Given that:-

x+ 2(x+1) + 3(x+2) =76

Solution:-

x+2x+2+3x+6=76

6x+8=76

6x=76-8

6x=68

x=68/6=11.33333...

So, the numbers are 11.33..,12.33..,13.33..

Hope it helps!