Three positive numbers form an increasing G.P | vedclass

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(B) Let the three positive numbers in $G.P.$ be $a, ar, ar^2$ where $r > 1$ for an increasing $G.P.$If the middle term is doubled,the numbers become $

Vedclass · Accepted Answer

$2 + \sqrt{3}$

Vedclass · Answer

(B) Let the three positive numbers in $G.P.$ be $a, ar, ar^2$ where $r > 1$ for an increasing $G.P.$If the middle term is doubled,the numbers become $a, 2ar, ar^2$.Since these are in $A.P.$,the middle term is the arithmetic mean of the other two:$2(2ar) = a + ar^2$$4ar = a(1 + r^2)$Since $a$ is a positive number,we can divide by $a$:$r^2 - 4r + 1 = 0$Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$r = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$Since the $G.P.$ is increasing,the common ratio $r$ must be greater than $1$.Thus,$r = 2 + \sqrt{3}$.

Three positive numbers form an increasing $G.P.$ If the middle term in this $G.P.$ is doubled,the new numbers are in $A.P.$ then the common ratio of the $G.P.$ is:

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