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 $b^2 = ac$.Therefore,$(9 + x)^2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(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 $b^2 = ac$.Therefore,$(9 + x)^2 = (3 + x)(21 + x)$.Expanding both sides:$81 + x^2 + 18x = 63 + 21x + 3x + x^2$$81 + 18x = 63 + 24x$$81 - 63 = 24x - 18x$$18 = 6x$$x = 3$.Verification: The numbers are $3+3=6$,$9+3=12$,and $21+3=24$. Since $6, 12, 24$ form a $G.P.$ with a common ratio of $2$,the value $x = 3$ is correct.

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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