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

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(B) Let $G_1, G_2, G_3, G_4, G_5$ be the five $G.M.s$ inserted between $a = 486$ and $b = 2/3$.The total number of terms in the sequence is $n = 5 + 2

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let $G_1, G_2, G_3, G_4, G_5$ be the five $G.M.s$ inserted between $a = 486$ and $b = 2/3$.The total number of terms in the sequence is $n = 5 + 2 = 7$.The $7^{th}$ term is given by $T_7 = ar^{7-1} = ar^6$.Substituting the values: $2/3 = 486 	imes r^6$.$r^6 = \frac{2}{3 	imes 486} = \frac{2}{1458} = \frac{1}{729} = (1/3)^6$.Thus,$r = 1/3$.The fourth $G.M.$ is the $5^{th}$ term of the sequence,$T_5 = ar^{5-1} = ar^4$.$T_5 = 486 	imes (1/3)^4 = 486 	imes \frac{1}{81} = 6$.

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

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