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(B) Let the number to be added be $x$.Then,the numbers $x + 2, x + 14, x + 62$ are in $G.P.$For three numbers $a, b, c$ to be in $G.P.$,the condition

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$2$

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(B) Let the number to be added be $x$.Then,the numbers $x + 2, x + 14, x + 62$ are in $G.P.$For three numbers $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.Therefore,$(x + 14)^2 = (x + 2)(x + 62)$.Expanding both sides: $x^2 + 28x + 196 = x^2 + 64x + 124$.Subtracting $x^2$ from both sides: $28x + 196 = 64x + 124$.Rearranging the terms: $196 - 124 = 64x - 28x$.$72 = 36x$.$x = 2$.Thus,adding $2$ to the given numbers results in $4, 16, 64$,which are in $G.P.$ as $16^2 = 4 	imes 64$ $(256 = 256)$.

The number which should be added to the numbers $2, 14, 62$ so that the resulting numbers may be in $G.P.$ is

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