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(A) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since $a - b, c - a, b - c$ are in $H.P.$,their reciprocals are in $A.P.$:$\frac{1}{c - a} =

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(A) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since $a - b, c - a, b - c$ are in $H.P.$,their reciprocals are in $A.P.$:$\frac{1}{c - a} = \frac{1}{2} \left( \frac{1}{a - b} + \frac{1}{b - c} \right)$$\frac{2}{c - a} = \frac{(b - c) + (a - b)}{(a - b)(b - c)} = \frac{a - c}{(a - b)(b - c)}$$2(a - b)(b - c) = -(a - c)^2$$2(ab - ac - b^2 + bc) = -(a^2 - 2ac + c^2)$Substituting $ac = b^2$:$2(ab - b^2 - b^2 + bc) = -(a^2 - 2b^2 + c^2)$$2ab - 4b^2 + 2bc = -a^2 + 2b^2 - c^2$$a^2 + 2ab + 2bc + c^2 = 6b^2$Since $b^2 = ac$,$a^2 + 2ab + 2bc + c^2 = 6ac$Alternatively,using the property $2(a-b)(b-c) = -(a-c)^2$:$2(ab - ac - b^2 + bc) = -(a^2 - 2ac + c^2)$$2ab - 2b^2 - 2ac + 2bc = -a^2 + 2ac - c^2$$a^2 + 2ab + 2bc + c^2 = 4ac + 2b^2 = 4b^2 + 2b^2 = 6b^2$This simplifies to $a + 4b + c = 0$.

If $a, b, c$ are in $G.P.$,and $a - b, c - a, b - c$ are in $H.P.$,then $a + 4b + c$ is equal to

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