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`        If-a-b-c-be-respectfully-the-pth-qth-rth-terms-of hp then sigma bc(q-r) is`
one year ago

Arun
23381 Points
```							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
```
one year ago
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