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$There are n-AM's between 1 and 31 such that 7^{th} {mean:(n-1)}^{th} mean = 5 : 9$ $then find the value of n$

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answer is B.

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Detailed Solution

$d = \frac{31 - 1}{n + 1} = \frac{30}{n + 1}$

$\begin{array}{l} According to the question \frac{a + 7 d}{a + (n - 1) d} = 5 / 9 \\ \Rightarrow 4 = (5 n-68) \frac{30}{n + 1} \Rightarrow 4 n + 4 = 150 n - 2040 \\ \Rightarrow n = 14 \end{array}$

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