Let $3$ geometric means $G_{1}, G_{2}, G_{3}$ are inserted between two positive number $a$ and $b$ such that $\frac{G_{3} - G_{2}}{G_{2} - G_{1}} = 2$ , then $\frac{b}{a}$ equal to

2

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16

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Solution

The correct option is B $2$
$F o r m u l a u s e d$
$T_{n} = a r^{n - 1}$
$G i v e n$
$a, G_{1}, G_{2}, G_{3}, b i n G . P$
$L e t r b e t h e c o m m o n r a t i o$
$G_{1} = a r$
$G_{2} = a r^{2}$
$G_{3} = a r^{3}$
$b = a r^{4}$
$G i v e n$
$\frac{G_{3} - G_{2}}{G_{2} - G_{1}} = 2$
$\Rightarrow \frac{a r^{3} - a r^{2}}{a r^{2} - a r} = 2$
$\Rightarrow \frac{a r^{2} (r - 1)}{a r (r - 1)} = 2$
$\Rightarrow r = 2$
$H e n c e$
$b = a r^{4}$
$= a {(2)}^{4}$
$= 16 a$
$H e n c e$
$\frac{b}{a} = \frac{16 a}{a} = 16$

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Let 3 geometric means G1,G2,G3 are inserted between two positive number a and b such that G3−G2G2−G1=2, then ba equal to

The correct option is B 2 Formulaused Tn=arn−1 Given a,G1,G2,G3,binG.P Letrbethecommonratio G1=ar G2=ar2 G3=ar3 b=ar4 Given G3−G2G2−G1=2 ⇒ar3−ar2ar2−ar=2 ⇒ar2(r−1)ar(r−1)=2 ⇒r=2 Hence b=ar4 =a(2)4 =16a Hence ba=16aa=16

Let $3$ geometric means $G_{1}, G_{2}, G_{3}$ are inserted between two positive number $a$ and $b$ such that $\frac{G_{3} - G_{2}}{G_{2} - G_{1}} = 2$ , then $\frac{b}{a}$ equal to