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(D) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Given that $b - c, c - a, a - b$ are in $H.P.$,we have $\frac{2}{c - a} = \frac{1}{b - c} +

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(D) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Given that $b - c, c - a, a - b$ are in $H.P.$,we have $\frac{2}{c - a} = \frac{1}{b - c} + \frac{1}{a - b}$.Simplifying the $H.P.$ condition: $\frac{2}{c - a} = \frac{a - b + b - c}{(b - c)(a - b)} = \frac{a - c}{(b - c)(a - b)}$.$2(b - c)(a - b) = -(c - a)^2$.$2(ab - b^2 - ac + bc) = -(c - a)^2$.Since $b^2 = ac$,we have $2(ab - ac - ac + bc) = -(c - a)^2$.$2(ab + bc - 2ac) = -(c - a)^2$.$2b(a + c) - 4ac = -(c^2 - 2ac + a^2)$.$2b(a + c) - 4b^2 = -c^2 + 2b^2 - a^2$.$a^2 + 2b(a + c) + c^2 = 6b^2$.Since $a+b+c = a+c+\sqrt{ac}$,the expression $a+b+c$ depends on $a, b,$ and $c$. Therefore,it is not independent of any of them.

If three unequal non-zero real numbers $a, b, c$ are in $G.P.$ and $b - c, c - a, a - b$ are in $H.P.$,then the value of $a + b + c$ is independent of

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