if `(1+3+5+7+....(2p-1))+(1+3+5+...+(2q-1)) =1+3+5+...+(2r -1),` then least possible value of `p+q+r `(Given `p>5`) is:

Consider the equation p = 5-2q - 3 Greatest sat of all possible values of p for q in R is

Consider the equation p = 5-2q - 3 Greatest sat of all possible values of p for q in R is

if 2^(p)+3^(q)=17 and 2^(p+2)-3^(q+1)=5 then find the value of p&q

The positive integers p,q and r, then possible value of r is 3(b)5(c)7(d)1

If p is an integer such that -5ltplt5 and q=3p-p^(3) , what is the least possible value of q?

If p = 3 , q = 2 and r = -1, find the values of (3p^(2)q+5r)/(2pq-3qr)

If 2p+3q+4r=15, the maximum value of p^(3)q^(5)r^(7) will be

If (1+3+5+...+p)+(1+3+5+...+q)=(1+3+5+...+r) where each set of parentheses contains the sum of consecutive odd integers as shown,the smallest possible value of p+q+r(where p>6) is 12 b.21 c.45 d.54