For x≥0, the least value of a, for which 51+x+51-x, a2, 25x+25-x are three consecutive terms of an A.P., is equal to:
(2)
Let 5x=t; t>0
A.M.≥G.M.
y2+5y-(2+a)=0; where y=t+1t≥2
Since a≥12