Ques no 7

Dear Student,
Let the three consecutive multiples of 9 be 9(x-1), 9x and 9(x+1). 
[ Note: we start with x-1 so that when the three numbers are added, -1 of first term gets cancelled out with the +1 of the third term and calculation becomes easy.] 

Their sum= 999
9(x-1)+9x+9(x+1) = 999 
9x-9+9x+9x+9 = 999
27x= 999
x= $\frac{999}{27}$= 37

1st number = 9(x-1) = 9(37-1)=9×36= 324
 2nd number = 9x= 9×37= 333
3rd number = 9(x+1)=9×38=342

Regards