If a (b1+c1), b(c1+a1), c(a1+b1) are in A.P. prove that a, b, c are in A.P.

Solution: As, these three values are in A.P.∴b(c1+a1)−a(b1+c1)=c(a1+b1)−b(c1+a1)⇒cb+ab−ba−ca=ac+bc−cb−ab⇒c1(b−a)+(abb2−a2)=a1(c−b)+(bcc2−b2)⇒c1(b−a)+ab(b−a)(b+a))=a1(c−b)+bc(c−b)(c+b))⇒(b−a)(c1+abb+a)=(c−b)(a1+cbc+b)⇒(b−a)(c1+a1+b1)=(c−b)(a1+b1+c1)⇒b−a=c−bSo, a,bandc are in AP.

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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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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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Text solutionVerified

Solution: As, these three values are in A.P.

∴ 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})

\Rightarrow \frac{b}{c} + \frac{b}{a} - \frac{a}{b} - \frac{a}{c} = \frac{c}{a} + \frac{c}{b} - \frac{b}{c} - \frac{b}{a}

\Rightarrow \frac{1}{c} (b - a) + (\frac{b ^{2} - a ^{2}}{a b}) = \frac{1}{a} (c - b) + (\frac{c ^{2} - b ^{2}}{b c})

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

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

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

\Rightarrow b - a = c - b

So,

a, b and c

are in AP.

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Question Text	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.
Answer Type	Text solution:1
Upvotes	150

If a (b1​+c1​), b(c1​+a1​), c(a1​+b1​) are in A.P. prove that a, b, c are in A.P.

Text solutionVerified

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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.