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# If $a\bigg(\large\frac{1}{b}+\frac{1}{c}\bigg)$$,\:b\bigg(\large\frac{1}{c}+\frac{1}{a}\bigg)$$,\:c\bigg(\large\frac{1}{a}+\frac{1}{b}\bigg)$ are in A.P., then prove that $a,b,c$ are in A.P.

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• If $a,b,c$ are in A.P. then $2b=a+c$
Given: $a\bigg(\large\frac{1}{b}+\frac{1}{c}\bigg)$$,\:b\bigg(\large\frac{1}{c}+\frac{1}{a}\bigg)$$,\:c\bigg(\large\frac{1}{a}+\frac{1}{b}\bigg)$ are in A.P.
$\Rightarrow\:2.b\bigg(\large\frac{1}{c}+\frac{1}{a}\bigg)$$=a\bigg(\large\frac{1}{b}+\frac{1}{c}\bigg)$$+c\bigg(\large\frac{1}{a}+\frac{1}{b}\bigg)$
Taking L.C.M. on both the sides and simplifying
$\Rightarrow\:\large\frac{2b(a+c)}{ac}=\frac{a^2(b+c)+c^2(b+a)}{abc}$
$\Rightarrow\:2b^2(a+c)=a^2(b+c)+c^2(b+a)$
$\Rightarrow\:2b^2(a+c)=a^2b+a^2c+c^2b+c^2a$
$\Rightarrow\:2b^2(a+c)=b(a^2+c^2)+ac(a+c)$
Adding and subtracting 2abc we get
$\Rightarrow\:2b^2(a+c)=b(a^2+c^2)+ac(a+c)+2abc-2abc$
$\Rightarrow\:2b^2(a+c)=b(a^2+c^2+2ac)+ac(a+c-2b)$
$\Rightarrow\:2b^2(a+c)=b(a+c)^2+ac(a+c-2b)$
$\Rightarrow\:2b^2(a+c)-b(a+c)^2=ac(a+c-2b)$
$\Rightarrow\:b(a+c)(2b-(a+c))=ac(a+c-2b)$
$\Rightarrow\:-b(a+c)(a+c-2b)=ac(a+c-2b)$
$\Rightarrow\:(a+c-2b)(ac+ab+bc)=0$
Since $ac,ab,bc$ are not zero $(ab+bc+ac)$ cannot be 0
$\Rightarrow\:a+c-2b=0$
$\Rightarrow\:2b=a+c$
$\Rightarrow\:a,b,c$ are in A.P.
Hence proved.
answered Mar 31, 2014