200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18…

For the given A.P.,
First term, a = 20
and common difference,d
\(= a_2 - a_1 = 19-20 = -1\)
Let a total of 200 logs be placed in n rows.
Thus, \(S_n = 200\)

By the sum of \(n^{th}\) term formula,
\(\Rightarrow S_n = {{n} \over {2}} [2a +(n -1)d]\)
\(\Rightarrow S_{12} = {{12} \over {2}} [2(20)+(n -1)(-1)]\)
\(\Rightarrow \) 400 = n (40- n+1)
\(\Rightarrow \) 400 = n (41-n)
\(\Rightarrow 400 = 41n-n^2\)
\(\Rightarrow n^2-41n + 400 = 0\)
\(\Rightarrow n^2-16n-25n+400 = 0\)
\(\Rightarrow \) n(n -16)-25(n -16) = 0
\(\Rightarrow\) (n -16)(n -25) = 0
Either (n -16) = 0 or n-25 = 0
\(\Rightarrow \) n = 16 or n = 25

By the nth term formula,
\(\Rightarrow a_n = a+(n-1)d\)
\(\Rightarrow a_{16} = 20+(16-1)(-1)\)
\(\Rightarrow a_{16} = 20-15\)
\(\Rightarrow a_{16} = 5\)

Similarly, the 25th term could be written as;
\(\Rightarrow a_{25} = 20+(25-1)(-1)\)
\(\Rightarrow a_{25} = 20-24\)
\(\Rightarrow a_{25}= -4 \)

It can be seen, the number of logs in 16th row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.