If $n$  arithmetic means $a_1,a_2,\cdots ,a_n$  are inserted between 50 and 200 and $n$  harmonic means $h_1,h_2,\cdots h_n$  are inserted between the same two numbers, then $a_2{h}_{n-1}$  is equal to $K$  then $\frac{K-9000}{500}=$

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

$a_2=50+2\left(\frac{150}{n+1}\right)=\frac{50\left(n+1\right)+300}{n+1}$

$50,h,h_2,\cdots 200\to H.P$

$\Rightarrow \frac{1}{50},\frac{1}{h_1},\frac{1}{h_2},\cdots \frac{1}{h_n},\frac{1}{200}\to A.P$

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

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

$a_2\times h_{n-1}=\frac{50\left(n+7\right)}{n+1}\times \frac{200\left(n+1\right)}{n+7}=10000,\frac{K-9000}{500}=\frac{1000}{500}=2$