The sum of three numbers in a GP is 105 and t | vedclass

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(B) Let the three numbers in $GP$ be $\frac{a}{r}, a, ar$.Given that their product is $8000$,so $\frac{a}{r} 	imes a 	imes ar = 8000$.This simplifie

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$5, 20, 80$

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(B) Let the three numbers in $GP$ be $\frac{a}{r}, a, ar$.Given that their product is $8000$,so $\frac{a}{r} 	imes a 	imes ar = 8000$.This simplifies to $a^3 = 8000$,which gives $a = 20$.Given that their sum is $105$,so $\frac{a}{r} + a + ar = 105$.Substituting $a = 20$,we get $\frac{20}{r} + 20 + 20r = 105$.Dividing by $5$,we get $\frac{4}{r} + 4 + 4r = 21$,which simplifies to $4r^2 - 17r + 4 = 0$.Solving the quadratic equation $4r^2 - 16r - r + 4 = 0$,we get $4r(r - 4) - 1(r - 4) = 0$.Thus,$(4r - 1)(r - 4) = 0$,giving $r = 4$ or $r = \frac{1}{4}$.If $r = 4$,the numbers are $\frac{20}{4}, 20, 20 	imes 4$,which are $5, 20, 80$.If $r = \frac{1}{4}$,the numbers are $\frac{20}{1/4}, 20, 20 	imes \frac{1}{4}$,which are $80, 20, 5$.In both cases,the set of numbers is ${5, 20, 80}$.

The sum of three numbers in a $GP$ is $105$ and their product is $8000$. Find the numbers.

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