If five G.M.s are inserted between 486 and 2/ | vedclass

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(B) Let the five geometric means be $G_1, G_2, G_3, G_4, G_5$ inserted between $a = 486$ and $b = 2/3$.The total number of terms in the resulting geom

Vedclass · Accepted Answer

$6$

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(B) Let the five geometric means be $G_1, G_2, G_3, G_4, G_5$ inserted between $a = 486$ and $b = 2/3$.The total number of terms in the resulting geometric progression is $n = 5 + 2 = 7$.The $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.Here,$T_7 = 2/3$,so $486 \cdot r^{7-1} = 2/3$.$r^6 = \frac{2}{3 \cdot 486} = \frac{2}{1458} = \frac{1}{729}$.Since $729 = 3^6$,we have $r^6 = (1/3)^6$,which gives $r = 1/3$.The fourth geometric mean is the fifth term of the progression,$T_5 = ar^{5-1} = ar^4$.$T_5 = 486 \cdot (1/3)^4 = 486 \cdot (1/81) = 6$.

If five $G.M.s$ are inserted between $486$ and $2/3$,then the fourth $G.M.$ will be:

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