Assersion (A):   If the angles of a convex polygon are in A.P, ${120}^o,{125}^o,{130}^o,\mathrm{......},$ then it has 16 sides.
Reason (R):  The sum of the angles of a polygon of n sides is $(n-2){180}^o.$ 

Let n be the number of sides 
$\frac{n}{2}\left[{240}^o+(n-1)5^o\right]=(n-2){180}^o$ 
$\Rightarrow 240n+5n^2-5n=360n-720$  
$\Rightarrow 5n^2-125n+720=0$ 
$\Rightarrow n^2-25n+144=0\Rightarrow n=9,16$ 
But $n=16,$ the largest angle is ${120}^o+15\times 5^o={195}^o>{180}^o$  
Not possible in a convex polygon
$\therefore n=9$