If n arithmetic mean are inserted between a and 100 such that the ratio of the first mean to the last mean is 1:7 and a+n=33, then the value of n is

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If $n$ arithmetic mean are inserted between $a$ and 100 such that the ratio of the first mean to the last mean is 1:7 and $a + n = 33,$ then the value of $n$ is

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Detailed Solution

Here $d = \frac{100 - a}{n + 1}$

Now, $A_{1} = a + d$ and, $A_{n} = 100 - d$ So, $\Rightarrow \frac{A_{1}}{A_{n}} = \frac{1}{7} \Rightarrow \frac{a + d}{100 - d} = \frac{1}{7} \Rightarrow 7 a + 8 d = 100$