Question

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Answer 1

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Answer 2

Answer:

pth term= a

qth term = b

rth term= c

To prove (q-r)bc +(r-p)ac+(p-q)ab = 0

Since a, b, c are the terms of H.P. hence 1/a, 1/b and 1/c are the corresponding terms of A.P.

x be the first term and 'd' be the common difference of the A.P.

→ 1/a =x+(p−1)d ...........(i)

1/b =x+(q−1)d ..................(ii)

1/c =x+(r−1)d ..................(iii)

(i) − (ii) gives 1/a −1/b = (p−q)d i.e. (b −a) = (p−q)d×ab ......(iv)

(ii) − (iii) gives 1/b −1/c = (q−r)d i.e. (c −b) = (p−r)d×bc ......(v)

(iii) − (i) gives 1/c −1/a = (r−p)d i.e. (a −c) = (r−p)d×ca ......(vi

Adding (iv), (v) and (vi), we get

(b −a) + (c −b) + (a −c)= (p−q)d×ab + (p−r)d×bc +(r−p)d×ca

or 0={ (p−q)ab + (p−r)bc +(r−p)ca }×d

Since d ≠ 0, we get

(p−q)ab + (p−r) bc +(r−p)ca =0