Number of ways of selecting 4 numbers. 
Number of ways of selecting 4 numbers from the set of first 20 Natural numbers, such that they are in A.P.

Umm, I did try solving it but couldn't come up with something concrete. I really don't want to evaluate each case separately, that'd be really cumbersome and not-so-smart way doing it, IMO. 

To get an arithmetic progression, of the form $a,a+d,a+2d,a+3d$, we need to choose $a$ and $d$, where $d \neq 0$.
If $a=1$, we need $3d \leq 19 \implies d \leq 6$. Hence, there are $6$ options.
In general, for a fixed $a$, $d$ has $\left \lfloor\dfrac{20-a}3 \right \rfloor$ options. Now sum up over all options to get the answer.

Hence, the number of ways is $$\sum_{a=1}^{17} \left \lfloor\dfrac{20-a}3 \right \rfloor = 57$$If you allow $d$ to be zero as well i.e. if you consider $a,a,a,a$ to be an arithmetic progression as well, then $$\sum_{a=1}^{17} \left \lfloor\dfrac{20-a}3 \right\rfloor  +20= 77$$

