The digits of a three digit number $N$ are in A.P.If sum of the digits is 15 and the number obtained by re-versing the digits of the number is 594 less than the original number, then $\frac{1000}{N-252}$ is equal to 

Let three digits of $N$ be: Hundred’s digit $= a – d$

ten’s digit $= a,$

and unit’s digit $= a+d$

$(a–d)+a+(a+d)=15 fi a=5$

and

$N=100(5-d)+50+(5+d)$                                         (1)

Number obtained by reversing digits of $N$ is

$N_1=100(5+d)+50+(5-d)$                                       (2)

Now,

 $\begin{array}{l} 594=N-N_1=100(-2d)+2d\\ \Rightarrow 2d=-6\Rightarrow d=-3\end{array}$

Thus, $N$ = 852 and

$\frac{1000}{N-252}=\frac{1000}{600}=\frac{5}{3}$