If x, y, z are in G.P then 1/(1+log x) , 1/(1+log y) and 1/(1+log z) will be in

If x, y, z are in GP, then y² = xz, so 2log y = log x + log z. Let a = log x, b = log y, c = log z. Then 2b = a + c. For the sequence 1/(1+a), 1/(1+b), 1/(1+c) to be in HP, their reciprocals (1+a), (1+b), (1+c) must be in AP. Since 2(1+b) = 2 + 2b = 2 + a + c = (1+a) + (1+c), the reciprocals are in AP, so the original sequence is in HP.