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Question:

If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in AP, prove that $a^{2}(b+c), b^{2}(c+a)$

$\mathbf{c}^{2}(\mathbf{a}+\mathbf{b})$ are in AP.

Solution:

To prove: $a^{2}(b+c), b^{2}(c+a), c^{2}(a+b)$ are in A.P.

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

Proof: $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P.

$\Rightarrow\left(\frac{a}{b}+\frac{a}{c}\right),\left(\frac{b}{c}+\frac{b}{a}\right),\left(\frac{c}{a}+\frac{c}{b}\right)$ are in A.P.

$\Rightarrow\left(\frac{a c+a b}{b c}\right),\left(\frac{a b+b c}{c a}\right),\left(\frac{c b+a c}{a b}\right)$ are in A.P.

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Multiplying the A.P. with (abc)

$\Rightarrow\left(\frac{a c+a b}{b c}\right)(a b c),\left(\frac{a b+b c}{c a}\right)(a b c),\left(\frac{c b+a c}{a b}\right)(a b c)$, are in A.P.

$\Rightarrow[(a c+a b)(a)],[(a b+b c)(b)],[(c b+a c)(c)]$ are in A.P.

$\Rightarrow\left[\left(a^{2} c+a^{2} b\right)\right],\left[a b^{2}+b^{2} c\right],\left[c^{2} b+a c^{2}\right]$ are in A.P.

On rearranging,

$\Rightarrow\left[a^{2}(b+c)\right],\left[b^{2}(c+a)\right],\left[c^{2}(a+b)\right]$ are in A.P.

Hence Proved