the digits of a certain number of 3 arithmetical digits are in arithmetical progression.Their sum is 15.And the number is 129 times the first digit.What is the number???

Question

by using the formula..

please...

Answer 1

n, n+1, n+2

3n+3=15
100n+10(n+1)+n+2 = 129n

n = 4
check
400 +50 +6 = 516 ? No

try
n, n+2, n+3
3n+5 = 15
n = 10/3 no

n, n+3, n+6
3n+9 = 15
n = 2
200 + 50 + 8 = 258 YES !
so number is
258