Three numbers are in G.P. such that their sum | vedclass

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Question

(A) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$.Given that their product is $1728$,so $(\frac{a}{r}) 	imes a 	imes (ar) = 1728$.$a^3 = 1

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$.Given that their product is $1728$,so $(\frac{a}{r}) 	imes a 	imes (ar) = 1728$.$a^3 = 1728$,which implies $a = \sqrt[3]{1728} = 12$.Given that their sum is $38$,so $\frac{a}{r} + a + ar = 38$.Substituting $a = 12$,we get $\frac{12}{r} + 12 + 12r = 38$.$\frac{12}{r} + 12r = 26$.Dividing by $2$,we get $\frac{6}{r} + 6r = 13$.$6 + 6r^2 = 13r$,which gives $6r^2 - 13r + 6 = 0$.Solving the quadratic equation $6r^2 - 9r - 4r + 6 = 0$,we get $3r(2r - 3) - 2(2r - 3) = 0$.$(3r - 2)(2r - 3) = 0$,so $r = \frac{2}{3}$ or $r = \frac{3}{2}$.If $r = \frac{3}{2}$,the numbers are $\frac{12}{3/2}, 12, 12 	imes \frac{3}{2} = 8, 12, 18$.If $r = \frac{2}{3}$,the numbers are $\frac{12}{2/3}, 12, 12 	imes \frac{2}{3} = 18, 12, 8$.In both cases,the numbers are $8, 12, 18$. The greatest number is $18$.

Three numbers are in $G.P.$ such that their sum is $38$ and their product is $1728$. The greatest number among them is

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