Let a,b  and c  be respectively the pth, qth  and rth  terms of a harmonic progression and Δ=111bqrbccaab  then numerical value of Δ  is

1a, 1b, 1c  are the pth, qth, rth  terms of an A.P.Let 1a=A+(p−1)D, 1b=A+(q−q)D ,And 1c=A+(r−1)D .Now Δ=abc|111pqrA+(p−1)DA+(q−1)DA+(r−1)D|Using R3→R3−(A−D)R1−DR2 , we getΔ=0