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(D) Let $x_1, x_2, \dots, x_m$ be the $m$ arithmetic means between $1$ and $31$.Then $1, x_1, x_2, \dots, x_m, 31$ forms an arithmetic progression wit

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$14$

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(D) Let $x_1, x_2, \dots, x_m$ be the $m$ arithmetic means between $1$ and $31$.Then $1, x_1, x_2, \dots, x_m, 31$ forms an arithmetic progression with $(m + 2)$ terms.Here,the first term $a = 1$ and the last term $T_{m+2} = 31$.Using the formula $T_n = a + (n - 1)d$,we have $31 = 1 + (m + 2 - 1)d$.$30 = (m + 1)d \implies d = \frac{30}{m + 1}$.The $k^{th}$ arithmetic mean is $x_k = a + kd$.Given $\frac{x_7}{x_{m-1}} = \frac{5}{9}$.Substituting the values: $\frac{1 + 7d}{1 + (m - 1)d} = \frac{5}{9}$.$9(1 + 7d) = 5(1 + (m - 1)d)$.$9 + 63d = 5 + 5(m - 1)d$.$4 = (5m - 5 - 63)d = (5m - 68)d$.$4 = (5m - 68) 	imes \frac{30}{m + 1}$.$4(m + 1) = 30(5m - 68)$.$4m + 4 = 150m - 2040$.$146m = 2044$.$m = \frac{2044}{146} = 14$.

If $m$ arithmetic means are inserted between $1$ and $31$ such that the ratio of the $7^{th}$ mean to the $(m - 1)^{th}$ mean is $5:9$,then the value of $m$ is:

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