Find 5 consecutive intergers such that the sum of the first and 3 times the fourth is equal to 39 less than twice the sum of the second,third and fifth .

Answer 1

Let the 5 consecutive integers be

x, x+1, x+2, x+3, & x+4

Based on the given information x + 3(x+3) = -39 + 2( x+1 + x+2 + x+4)

Collecting terms 4x + 9 = 6x - 25

or -2x = -34

or x = 17

So the digits are 17,18,19,20, & 21