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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$5, 20, 80$

Vedclass · Answer

(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.

Similar Questions

The sum of $1 + 3 + 5 + 7 + \dots$ up to $n$ terms is:

The sum to $n$ terms of $(2n - 1) + 2(2n - 3) + 3(2n - 5) + .....$ is

If $x \in (0, \frac{\pi}{4})$,then the expression $\frac{\cos x}{\sin^2 x(\cos x - \sin x)}$ cannot take the value

If $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-

If $\alpha, \beta$ are the roots of the equation $x^2 - 3x + a = 0$ and $\gamma, \delta$ are the roots of the equation $x^2 - 12x + b = 0$,and $\alpha, \beta, \gamma, \delta$ form an increasing $G.P.$,then $(a, b) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module