If 51+x+51−x,a2 and 25x+25−x  are three consecutive terms of an A.P., then the values of a are given by

Since, 51+x+51−x,a2,25x+25−x are in A.P., we have  2a2=51+x+51−x+25x+25−xNow put 5x = t so that t > 0, we then havea=5t+5t+t2+1t2=t2+1t2+5t+1tor   a=t−1t2+2+5t−1t2+2=t−1t2+5t−1t2+12≥12Thus, values of a are given by the inequality a ≥ 12.