If (a - b), (b - c), (c - a) are in G.P., then prove that: (a+b+c )^2=3 (ab+bc+ca )

(a - b), (b - c), (c - a) are in G.P. (b - c)^2 = (a - b) (c - a) b^2 + c^2 - 2bc = ac - a^2 - bc + ab b^2 + c^2 + a^2 = ac + bc + ab (i) Now, (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca =ac+bc+ab+2ab+2bc+2bc+2ca [Using equation (1)] =3ab+3bc+3ca (a+b+c)^2=3(ab+bc+ca)

If (a - b), (b - c), (c - a) are in G.P., then prove that: (a+b+c…

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If (a - b), (b - c), (c - a) are in G.P., then prove that:
$\big(\text{a}+\text{b}+\text{c}\big)^2=3\big(\text{ab}+\text{bc}+\text{ca}\big)$

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$(\text{a} - \text{b}), (\text{b} - \text{c}), (\text{c} - \text{a}) \text{ are in G.P.}$
$(\text{b} - \text{c})^2 = (\text{a} - \text{b}) (\text{c} - \text{a})$
$\text{b}^2 + \text{c}^2 - 2\text{bc} = \text{ac} - \text{a}^2 - \text{bc} + \text{ab}$
$\text{b}^2 + \text{c}^2 + \text{a}^2 = \text{ac} + \text{bc} + \text{ab}\cdots(\text{i})$
Now,
$(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}$
$=\text{ac}+\text{bc}+\text{ab}+2\text{ab}+2\text{bc}+2\text{bc}+2\text{ca}$ [Using equation (1)]
$=3\text{ab}+3\text{bc}+3\text{ca}$
$(\text{a}+\text{b}+\text{c})^2=3(\text{ab}+\text{bc}+\text{ca})$

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