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  • If a 1 / b + 1 / c b 1 / c + 1 / a c 1/ a+ 1/ b are in A.P., prove that a, b, c are in A.P.

16.  If  a (\frac{1}{b}+\frac{1}{c}) , b ( \frac{1}{c}+\frac{1}{a}) , c ( \frac{1}{a}+ \frac{1}{b})  are in A.P., prove that a, b, c are in A.P.

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Given:a (\frac{1}{b}+\frac{1}{c}) , b ( \frac{1}{c}+\frac{1}{a}) , c ( \frac{1}{a}+ \frac{1}{b}) are in A.P.

\therefore \, \, \, b ( \frac{1}{c}+\frac{1}{a})-a (\frac{1}{b}+\frac{1}{c}) = c ( \frac{1}{a}+ \frac{1}{b})- b ( \frac{1}{c}+\frac{1}{a})

 \therefore \, \, \, ( \frac{b(a+c)}{ac})- (\frac{a(c+b)}{bc}) = ( \frac{c(b+a)}{ab})- ( \frac{b(a+c)}{ac})

\therefore \, \, \, ( \frac{ab^2+b^2c-a^2c-a^2b}{abc}) = ( \frac{c^2b+c^2a-b^2a-b^2c}{abc})

\Rightarrow \, \, \, ab^2+b^2c-a^2c-a^2b= c^2b+c^2a-b^2a-b^2c

\Rightarrow \, \, \, ab(b-a)+c(b^2-a^2)= a(c^2-b^2)+bc(c-b)

\Rightarrow \, \, \, ab(b-a)+c(b-a)(b+a)= a(c-b)(c+b)+bc(c-b)

\Rightarrow \, \, \, (b-a)(ab+c(b+a))= (c-b)(a(c+b)+bc)

\Rightarrow \, \, \, (b-a)(ab+cb+ac)= (c-b)(ac+ab+bc)

\Rightarrow \, \, \, (b-a)= (c-b)

Thus, a,b,c  are in AP.

Posted by

seema garhwal

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