Question

2. Find four consecutive integers such that six times the sum of the first and third is 12

greater than 9 times the fourth.

Answer 1

9

Explanation:

a,a+1,a+2,a+3,...

6(a+(a+2))=12+9(a+3)

6(2a+2)=12+9a+27

12a+12=12+9a+27

12a-9a=27

3a=27

a=9