If x is added to each of the numbers 3, 9, 21 | vedclass

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(A) Given that $3 + x, 9 + x, 21 + x$ are in $G.P.$For three numbers $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.Applying this condition: $(

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$3$

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(A) Given that $3 + x, 9 + x, 21 + x$ are in $G.P.$For three numbers $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.Applying this condition: $(9 + x)^2 = (3 + x)(21 + x)$Expanding both sides: $81 + x^2 + 18x = 63 + 21x + 3x + x^2$Simplifying the equation: $81 + 18x = 63 + 24x$Rearranging terms: $81 - 63 = 24x - 18x$$18 = 6x$$x = 3$Verification: If $x = 3$,the numbers are $3+3=6, 9+3=12, 21+3=24$. Since $12/6 = 2$ and $24/12 = 2$,the common ratio is constant,confirming they are in $G.P.$

If $x$ is added to each of the numbers $3, 9, 21$ so that the resulting numbers are in $G.P.$,then the value of $x$ is:

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