let the number be 'x' and the digits be

(a-d) at unit's place,

a at 10th place, &

(a+d) at 100th place

this is true since we know that their are in A.P.

then we can write for the number

x = (a-d) + 10*a + 100*(a + d)

the sum of the digits is 18

a + d + a + a - d = 18

3*a = 18

a = 6

the number obtained by reversing digits is: a + d + 10*a + 100*(a - d) = 111*a - 99*d is 594 less than original number

(a - d) + 10*a + 100*(a + d) - 594 = a + d + 10*a + 100*(a - d)

111*a + 99*d - 594 = 111*a - 99*d

198*d = 594

d = 3

so the original number has digits: 3, 6, 9 thus the number is 963

indeed

3 + 6 + 9 =19

963 = 594 + 369

the original number is 936.