If 25, x - 6,and x - 12 are consecutive terms | vedclass

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(C) Given that $25, x - 6$,and $x - 12$ are in a geometric progression $(GP)$.For three terms $a, b, c$ to be in $GP$,the condition is $b^2 = ac$.Subs

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Given that $25, x - 6$,and $x - 12$ are in a geometric progression $(GP)$.For three terms $a, b, c$ to be in $GP$,the condition is $b^2 = ac$.Substituting the values: $(x - 6)^2 = 25(x - 12)$.Expanding the equation: $x^2 - 12x + 36 = 25x - 300$.Rearranging into a quadratic equation: $x^2 - 37x + 336 = 0$.Factoring the quadratic: $(x - 16)(x - 21) = 0$.Thus,$x = 16$ or $x = 21$.Comparing with the given options,the correct value is $x = 16$.

If $25, x - 6$,and $x - 12$ are consecutive terms of a geometric progression,then $x = \dots$

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