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If $a, b, c$ are in G.P., $a-b, c-a, b-c$ are in H.P., then $a+4 b+c$ is equal to
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$b^2=a c$ and $\frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c}=\frac{a-c}{(a-b)(b-c)}$
$\Rightarrow \quad 2(a-b)(b-c)=-(a-c)^2$
$\Rightarrow \quad-(a-c)^2=2\left(a b-2 b^2+b c\right)=2 b\{a-2 \sqrt{a c}+c\}$
$\Rightarrow \quad-\{(\sqrt{a}-\sqrt{c})(\sqrt{a}+\sqrt{c})\}^2=2 b(\sqrt{a}-\sqrt{c})^2$
$\Rightarrow 2 b=-\{a+2 \sqrt{a c}+c\}=-a-2 b-c$ or $a+4 b+c=0$.
$b^2=a c$ and $\frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c}=\frac{a-c}{(a-b)(b-c)}$
$\Rightarrow \quad 2(a-b)(b-c)=-(a-c)^2$
$\Rightarrow \quad-(a-c)^2=2\left(a b-2 b^2+b c\right)=2 b\{a-2 \sqrt{a c}+c\}$
$\Rightarrow \quad-\{(\sqrt{a}-\sqrt{c})(\sqrt{a}+\sqrt{c})\}^2=2 b(\sqrt{a}-\sqrt{c})^2$
$\Rightarrow 2 b=-\{a+2 \sqrt{a c}+c\}=-a-2 b-c$ or $a+4 b+c=0$.
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