$200$ 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 (see Figure). In how many rows are the $200$ logs placed and how many logs are in the top row?

(N/A) It can be observed that the numbers of logs in rows form an $A.P.$
$20, 19, 18, \dots$
For this $A.P.$
$a = 20$
$d = a_2 - a_1 = 19 - 20 = -1$
Let a total of $200$ logs be placed in $n$ rows.
$S_n = 200$
Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
$200 = \frac{n}{2}[2(20) + (n - 1)(-1)]$
$400 = n(40 - n + 1)$
$400 = n(41 - n)$
$400 = 41n - n^2$
$n^2 - 41n + 400 = 0$
$n^2 - 16n - 25n + 400 = 0$
$n(n - 16) - 25(n - 16) = 0$
$(n - 16)(n - 25) = 0$
So,$n = 16$ or $n = 25$.
Now,find the number of logs in the $n^{th}$ row using $a_n = a + (n - 1)d$:
For $n = 16$:
$a_{16} = 20 + (16 - 1)(-1) = 20 - 15 = 5$
For $n = 25$:
$a_{25} = 20 + (25 - 1)(-1) = 20 - 24 = -4$
Since the number of logs cannot be negative,$n = 25$ is rejected.
Therefore,the $200$ logs are placed in $16$ rows,and there are $5$ logs in the top row.