Question

the sum of three consecutive multiples of 9 os 999.find the number.

Answer 1

9(n-1) × 9n × 9(n+1) = 999
(n-1)×(n+1)= 999/9×9×9
n^2 - 1 = 111/81
n^2 = 37/27 + 1
= 64/27
n = √64/27
= 8/3 × √1/3
8/3 × 3^-2
the numbers are
n-1 = (8/3 × 3^-2) - 1
n = 8/3 × 3^-2
n +1 = (8/3 × 3^-2) +1

the answer I have got is in this form only, I was expecting real numbers .
I hope that it helped ☺

Answer 2

solution :-

Given:-
The sum of three consecutive multiples of 9 is 378.

Let the first number be x.

second number= x +9

Third number = x+9+9 = x+18

Now

x + x+ 9 + x+18 = 999
=>3x + 27 = 999
=>3x = 999 -27
=>x = 972/3
=>x = 324

first number = 324

second number = 324+9 = 333

Third number = 324+18 = 342


Thanks