Q13/039) Show that 2^(1/2) 3^(1/2) and 5^(1/2) cannot be 3 terms of an AP

We know 2^(1/2) < 3^(1/2) < 5^(1/2)

Without loss of generality we can assume that 2^(1/2) is the 1^st term and < 3^(1/2) is the pth and 5^(1/2) is the qth term

If r is difference

So 2^(1/2) + (p-1) r = 3^(1/2)

2^(1/2) + (q-1) r = 5^(1/2)

Subtract to get (q-p) r = 5^(1/2) – 3^(1/2)

Or (q-p) r + 3^(1/2) = 5^(1/2)

Square both sides (q-p)^2 r + 3+ 2 sqrt(3)(q-p) = 5

Or sqrt(3) = (2-(q-p)^2)/2((q-p))

Right hand side is rational and LHS is irrational hence a contradiction

Hence proved