If n arithmetic means a_1, a_2, dots, a_n are | vedclass

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Question

(A) Let $50, a_1, a_2, \dots, a_n, 100$ be in an arithmetic progression $(A.P.)$.The common difference $d = \frac{100 - 50}{n + 1} = \frac{50}{n + 1}$

Vedclass · Accepted Answer

$5000$

Vedclass · Answer

(A) Let $50, a_1, a_2, \dots, a_n, 100$ be in an arithmetic progression $(A.P.)$.The common difference $d = \frac{100 - 50}{n + 1} = \frac{50}{n + 1}$.Then $a_2 = 50 + 2d = 50 + \frac{100}{n + 1} = \frac{50n + 50 + 100}{n + 1} = \frac{50(n + 3)}{n + 1}$.Let $50, h_1, h_2, \dots, h_n, 100$ be in a harmonic progression $(H.P.)$.Then $\frac{1}{50}, \frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}, \frac{1}{100}$ are in an $A.P.$The common difference $d' = \frac{\frac{1}{100} - \frac{1}{50}}{n + 1} = \frac{-1}{100(n + 1)}$.The $(n-1)$-th harmonic mean $h_{n-1}$ corresponds to the $(n-1+1)$-th term in the $A.P.$ of reciprocals,which is the $n$-th term.$\frac{1}{h_{n-1}} = \frac{1}{50} + (n-1)d' = \frac{1}{50} - \frac{n-1}{100(n + 1)} = \frac{2(n + 1) - (n - 1)}{100(n + 1)} = \frac{2n + 2 - n + 1}{100(n + 1)} = \frac{n + 3}{100(n + 1)}$.Thus,$h_{n-1} = \frac{100(n + 1)}{n + 3}$.Finally,$a_2 h_{n-1} = \left( \frac{50(n + 3)}{n + 1} \right) \cdot \left( \frac{100(n + 1)}{n + 3} \right) = 50 \cdot 100 = 5000$.

If $n$ arithmetic means $a_1, a_2, \dots, a_n$ are inserted between $50$ and $100$ and $n$ harmonic means $h_1, h_2, \dots, h_n$ are inserted between the same two numbers,then $a_2 h_{n-1}$ is equal to

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