The sum of three numbers in a Geometric Progr | vedclass

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Question

(A) Let the three numbers in $GP$ be $\frac{a}{r}, a, ar$.The product of these numbers is $\left( \frac{a}{r} ight) 	imes a 	imes (ar) = a^3 = 172

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) Let the three numbers in $GP$ be $\frac{a}{r}, a, ar$.The product of these numbers is $\left( \frac{a}{r} ight) 	imes a 	imes (ar) = a^3 = 1728$.Since $1728 = 12^3$,we have $a = 12$.The sum of the numbers is $\frac{a}{r} + a + ar = 38$.Substituting $a = 12$,we get $\frac{12}{r} + 12 + 12r = 38$.$\frac{12}{r} + 12r = 26$.Dividing by $2$,we get $\frac{6}{r} + 6r = 13$,which simplifies to $6r^2 - 13r + 6 = 0$.Factoring the quadratic equation: $6r^2 - 9r - 4r + 6 = 0 \implies 3r(2r - 3) - 2(2r - 3) = 0$.$(3r - 2)(2r - 3) = 0$,so $r = \frac{2}{3}$ or $r = \frac{3}{2}$.If $r = \frac{3}{2}$,the numbers are $\frac{12}{3/2}, 12, 12(\frac{3}{2}) \implies 8, 12, 18$.The largest number is $18$.

The sum of three numbers in a Geometric Progression $(GP)$ is $38$ and their product is $1728$. The largest of these numbers is.......

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