Grade 11IIT JEE Entrance Exam

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

Sudhanva G V

7 Years agoGrade 11

1 Answer

Arun

7 Years ago

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