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# if a2, b2,c2  are in A.P. then  prove that..a/b+c, b/c+a, c/a+b

dinesh
22 Points
6 years ago
it is very easy
use b2-a2=c2-b2

mycroft holmes
272 Points
6 years ago
If a2, b2, c2 are in A.P., so are a2+ab+bc+ca,  b2+ab+bc+ca, and  c2+ab+bc+ca

i.e. (a+b)(c+a), (b+c)(a+b), (c+a)(b+c) are in A.P.

Multiplying the three terms by (a+b+c)/ (a+b)(b+c)(c+a), we have

(a+b+c)/(b+c), (a+b+c)/c+a, (a+b+c)/a+b are in A.P. Now subtracting 1 from each term, we have

1/b+c, 1/c+a, 1/a+b are in A.P.
mycroft holmes
272 Points
6 years ago
If a2, b2, c2 are in A.P., so are a2+ab+bc+ca,  b2+ab+bc+ca, and  c2+ab+bc+ca

i.e. (a+b)(c+a), (b+c)(a+b), (c+a)(b+c) are in A.P.

Multiplying the three terms by (a+b+c)/ (a+b)(b+c)(c+a), we have

(a+b+c)/(b+c), (a+b+c)/c+a, (a+b+c)/a+b are in A.P. Now subtracting 1 from each term, we have

1/b+c, 1/c+a, 1/a+b are in A.P.