If there are 5 geometric means between 486 an | vedclass

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(A) Let the $5$ geometric means be $G_1, G_2, G_3, G_4, G_5$ between $a = 486$ and $b = \frac{2}{3}$.Then $486, G_1, G_2, G_3, G_4, G_5, \frac{2}{3}$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Let the $5$ geometric means be $G_1, G_2, G_3, G_4, G_5$ between $a = 486$ and $b = \frac{2}{3}$.Then $486, G_1, G_2, G_3, G_4, G_5, \frac{2}{3}$ are in a Geometric Progression $(GP)$.Here,the total number of terms $n = 5 + 2 = 7$.The $7^{th}$ term is $ar^{7-1} = ar^6 = \frac{2}{3}$.Substituting $a = 486$,we get $486 	imes r^6 = \frac{2}{3}$.$r^6 = \frac{2}{3 	imes 486} = \frac{2}{1458} = \frac{1}{729}$.Since $729 = 3^6$,we have $r^6 = (\frac{1}{3})^6$,so $r = \frac{1}{3}$ (taking the positive common ratio).The $4^{th}$ geometric mean is $G_4$,which is the $5^{th}$ term of the $GP$.$G_4 = ar^4 = 486 	imes (\frac{1}{3})^4$.$G_4 = 486 	imes \frac{1}{81} = 6$.

If there are $5$ geometric means between $486$ and $\frac{2}{3}$,what is the $4^{th}$ geometric mean?

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