If m arithmetic means (A.Ms) and three geomet | vedclass

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(A) Let the $m$ arithmetic means be $A_1, A_2, \dots, A_m$ between $3$ and $243$.The common difference $d$ is given by $d = \frac{243 - 3}{m + 1} = \f

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$39$

Vedclass · Answer

(A) Let the $m$ arithmetic means be $A_1, A_2, \dots, A_m$ between $3$ and $243$.The common difference $d$ is given by $d = \frac{243 - 3}{m + 1} = \frac{240}{m + 1}$.The $4^{	ext{th}}$ arithmetic mean is $A_4 = a + 4d = 3 + 4\left(\frac{240}{m + 1}ight)$.Let the three geometric means be $G_1, G_2, G_3$ between $3$ and $243$.The common ratio $r$ is given by $r = \left(\frac{243}{3}ight)^{\frac{1}{3 + 1}} = (81)^{1/4} = 3$.The $2^{	ext{nd}}$ geometric mean is $G_2 = ar^2 = 3(3)^2 = 27$.Given $A_4 = G_2$,we have:$3 + \frac{960}{m + 1} = 27$$\frac{960}{m + 1} = 24$$m + 1 = \frac{960}{24} = 40$$m = 39$.

If $m$ arithmetic means $(A.Ms)$ and three geometric means $(G.Ms)$ are inserted between $3$ and $243$ such that the $4^{\text{th}}$ $A.M.$ is equal to the $2^{\text{nd}}$ $G.M.$,then $m$ is equal to:

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