Question

Find four consecutive integers such that three times the sum of the first two integers exceeds the sum of the last two by 70.

Answer 1

we know the difference of consecutive integers is 1 so we can achieve every next integer by adding 1 in the previous integer .

let us start with assuming the four consecutive integers as :

x , x+1 , x+2 , x+3

3 times ( sum of first two ) = sum of last two + 70

3 ( x+ x+1) = x+2+x+3 + 70

3( 2x +1 ) = 2x + 75

6x + 3 = 2x + 75

6x-2x = 75 - 3

4x = 72

x= 72/4 = 18

so first integer is x= 18 next is 19 next 20 and next 21

Answer : 18 ,19,20,21