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

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Question

(A) Let the numbers be $A = 50$ and $B = 100$.For arithmetic means,the common difference $d = \frac{B-A}{n+1} = \frac{50}{n+1}$.Then $a_2 = A + 2d = 5

Vedclass · Accepted Answer

$5000$

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(A) Let the numbers be $A = 50$ and $B = 100$.For arithmetic means,the common difference $d = \frac{B-A}{n+1} = \frac{50}{n+1}$.Then $a_2 = A + 2d = 50 + \frac{100}{n+1} = \frac{50n + 50 + 100}{n+1} = \frac{50(n+3)}{n+1}$.For harmonic means,$\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$ are in $A$.$P$. with first term $\frac{1}{A} = \frac{1}{50}$ and last term $\frac{1}{B} = \frac{1}{100}$.The common difference $d' = \frac{\frac{1}{100} - \frac{1}{50}}{n+1} = \frac{-1}{100(n+1)}$.The term $\frac{1}{h_{n-1}}$ is the $(n-1)$-th term of the $A$.$P$. of reciprocals,which is $\frac{1}{A} + (n-1)d' = \frac{1}{50} - \frac{n-1}{100(n+1)}$.$\frac{1}{h_{n-1}} = \frac{2(n+1) - (n-1)}{100(n+1)} = \frac{2n+2-n+1}{100(n+1)} = \frac{n+3}{100(n+1)}$.Thus,$h_{n-1} = \frac{100(n+1)}{n+3}$.Finally,$a_2 h_{n-1} = \frac{50(n+3)}{n+1} 	imes \frac{100(n+1)}{n+3} = 5000$.

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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