200 logs are stacked in the following manner; 20 logs in the bottom row, 19 in the next row, 18 in the next to it and so on. Then the number of logs in the top row is

Series is 20, 19, 18, ….

$\begin{array}{l}d=19-20=-1\\ a=20\end{array}$

$\begin{array}{l}{S}_n=\frac{n}{2}[2a+(n-1)d]\\ \Rightarrow 200=\frac{n}{2}=(2\times 20+(n-1)(-1))\\ \Rightarrow 200=\frac{n}{2}(40-n+1)\\ \Rightarrow 400=40n-n^2+n\\ \Rightarrow n^2-41n+400=0\\ n^2-25n-16n+400=0\\ \Rightarrow n(n-25)-16(n-25)=0\\ (n-25)(n-16)=0\end{array}$

$n=25or16$

$\begin{array}{l}{T}_{16}=a+15d=20-15=5\\ T_{25}=a+24d=20-24=-4\end{array}$

Since $T_{25}$ is not possible

$\therefore$ n = 16  and 5 logs are placed in the top row.