If n arithmetic means a1,a2,⋯,an are inserted between 50 and 200 and n harmonic means h1,h2,⋯hn are inserted between the same two numbers, then a2hn−1 is equal to K then K−9000500=

50,a1,a2,⋯,200→A.P d=200−50n+1=150n+1 a2=50+2(150n+1)=50(n+1)+300n+1 50,h,h2,⋯200→H.P ⇒150,1h1,1h2,⋯1hn,1200→A.P d=1200−150n+1=−3200(n+1) ∴1hn−1=1200−2(−3200(n+1))=n+1+6200(n+1)⇒hn−1=200(n+1)n+7 a2×hn−1=50(n+7)n+1×200(n+1)n+7=10000, K−9000500=1000500=2

If n arithmetic means a1,a2,⋯,an are inserted between 50 and 200 and n harmonic means h1,h2,⋯hn are inserted between the same two numbers, then a2hn−1 is equal to K then K−9000500=

50,a1,a2,⋯,200→A.P d=200−50n+1=150n+1 a2=50+2(150n+1)=50(n+1)+300n+1 50,h,h2,⋯200→H.P ⇒150,1h1,1h2,⋯1hn,1200→A.P d=1200−150n+1=−3200(n+1) ∴1hn−1=1200−2(−3200(n+1))=n+1+6200(n+1)⇒hn−1=200(n+1)n+7 a2×hn−1=50(n+7)n+1×200(n+1)n+7=10000, K−9000500=1000500=2

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If $n$ arithmetic means $a_{1}, a_{2}, \dots, a_{n}$ are inserted between 50 and 200 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 $K$ then $\frac{K - 9000}{500} =$

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Detailed Solution

50, a_{1}, a_{2}, \dots, 200 \to A . P

$d = \frac{200 - 50}{n + 1} = \frac{150}{n + 1}$

$a_{2} = 50 + 2 (\frac{150}{n + 1}) = \frac{50 (n + 1) + 300}{n + 1}$

$50, h, h_{2}, \dots 200 \to H . P$

$\Rightarrow \frac{1}{50}, \frac{1}{h_{1}}, \frac{1}{h_{2}}, \dots \frac{1}{h_{n}}, \frac{1}{200} \to A . P$

$d = \frac{\frac{1}{200} - \frac{1}{50}}{n + 1} = \frac{- 3}{200 (n + 1)}$

$∴ \frac{1}{h_{n - 1}} = \frac{1}{200} - 2 (\frac{- 3}{200 (n + 1)}) = \frac{n + 1 + 6}{200 (n + 1)} \Rightarrow h_{n - 1} = \frac{200 (n + 1)}{n + 7}$