If n arithmetic means a_1,a_2,......a_n are i | vedclass

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Question

$50, a_{1}, a_{2}, \dots, a_{n}, 100$ are in $A.P.$

$\mathrm{a}_{2}=50+2 \mathrm{d} ;$ where $\mathrm{d}=\frac{50}{\mathrm{n}+1}$

$a_{2}=\left(5

Vedclass · Accepted Answer

$5000$

Vedclass · Answer

$50, a_{1}, a_{2}, \dots, a_{n}, 100$ are in $A.P.$

$\mathrm{a}_{2}=50+2 \mathrm{d} ;$ where $\mathrm{d}=\frac{50}{\mathrm{n}+1}$

$a_{2}=\left(50+\frac{100}{n+1}\right)=\frac{50 n+150}{n+1}=\frac{50(n+3)}{(n+1)}$

$50, \mathrm{h}_{1}, \mathrm{h}_{2}, \ldots \ldots . ., \mathrm{h}_{\mathrm{n}}, 100$ are in $H.P.$

$\frac{1}{h_{n-1}}=\frac{1}{50}+(n-1) d^{\prime};$

where $\mathrm{d}^{\prime}=\frac{\left(\frac{1}{100}-\frac{1}{50}\right)}{\mathrm{n}+1}=\frac{-1}{100(\mathrm{n}+1)}$

${\frac{1}{h_{n-1}}=\frac{1}{50}+(n-1) \cdot \frac{(-1)}{100(n+1)}} $

${\frac{1}{h_{n-1}}=\frac{2(n+1)-(n-1)}{100(n+1)} \Rightarrow h_{n-1}=\frac{100(n+1)}{(n+3)}} $

${a_{2} h_{n-1}=5000}.$

If $n$ arithmetic means $a_1,a_2,......a_n$ are inserted between $50$ and $100$ and $n$ harmonic means $h_1$ , $h_2$ , ...... $h_n$ are inserted between the same two numbers, then $a_2h_{n-1}$ is equal to

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