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(C) Let the first term of the arithmetic progression be $A$ and the common difference be $D$.Then,the terms are given by:$1/a = A + (p - 1)D \dots (1)

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$0$

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(C) Let the first term of the arithmetic progression be $A$ and the common difference be $D$.Then,the terms are given by:$1/a = A + (p - 1)D \dots (1)$$1/b = A + (q - 1)D \dots (2)$$1/c = A + (r - 1)D \dots (3)$Subtracting $(2)$ from $(1)$: $1/a - 1/b = (p - q)D \implies (b - a)/ab = (p - q)D \implies (a - b)/ab = -(p - q)D$Similarly,$(b - c)/bc = -(q - r)D$ and $(c - a)/ca = -(r - p)D$.Multiplying by $ab$,$bc$,and $ca$ respectively:$ab(p - q) = (a - b)/D$$bc(q - r) = (b - c)/D$$ca(r - p) = (c - a)/D$Adding these equations:$ab(p - q) + bc(q - r) + ca(r - p) = \frac{1}{D} (a - b + b - c + c - a) = \frac{1}{D} (0) = 0$.

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an arithmetic progression are $1/a$,$1/b$,and $1/c$ respectively,then $ab(p - q) + bc(q - r) + ca(r - p) = \dots$

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