Question

Iif b^(2)+c^(2) c^(2)+a^(2) a^(2)+b^(2) are in A.P. show that (1)/(b+c) (1)/(c+a) (1)/(a+b) are in A.P.

Answer 1

Answer:

a²,b²,c² are in A.P. adding a common term(ab+bc+ca) to each term could be in A.P.

So a²+ab+bc+ca,b²+ab+bc+ca,c²+ab+bc+ca are in A.P.

=>a(a+b)+c(a+b),b(b+a)+c(b+a),c²+ca+ab+bc are in A.P.

=>(a+b)(c+a),(b+c)(a+b),c(c+a)+b(c+a) are in A.P

=>(a+b)(c+a),(b+c)(a+b),(c+a)(b+c) are in A.P. , then dividing the (a+b)(b+c)(c+a) to all the three gives (a+b)(c+a)/(a+b)(b+c)(c+a),

(b+c)(a+b)/(a+b)(b+c)(c+a),(c+a)(b+c)/(a+b)(b+c)(c+a)

=>1/(b+c),1/(c+a),1/(a+b) are in A.P. proved

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