Set P consists of the first n positive multiples of 3 and set Q consis

Given that
P={3,6,9,...….3n}
Q={5,10,......5m}
Sum of a series with k terms , first term=a and last term=b is
Sum= k/2*(a+b))
so Sum of P= n/2*(3+3n)= 3n/2*(n+1) here n or n+1 is even and may be 2,4,6,----
so Sum of P =odd or even will depend on value of n or n+1
ex- if n is even and n>2 then Sum will always be even and if n=even but n=2 the sum= odd always
Similarly for Sum of Q= m/2*(5+5m)= 5m/2*(m+1) will depend on actual value of m or m+1
Stat 1--m=odd & n=even
Sum of P may be even or odd depending on actual value of 'n'
Sum of Q may be even or odd depending on actual value of 'm+1'
So R-S may be odd or even ----- not sufficient
Stat 2----m=4x+3 so m+1 =4(x+1) a multiple of 4 so Sum of Q= always even
but as n=2x it may be multiple of 4 or not So sum of P= even/odd
So R-S= even/odd-even=may or may be odd---- not sufficient
Stat1+2--- also doesn't give actual value of n and m so not sufficient
Ans ---E