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

Hint: In this particular type of question use the concept of arithmetic progression series as the number of logs is decreased by one so it is in A.P with common difference (-1) and first term 20, and the sum of the A.P series is given as, ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
Total number of logs = 200.
Number of logs in the first row = 20
Number of logs in the second row = 19
Number of logs in the third row = 18
And so on...
So, it is written as in series form
20, 19, 18, .........
Now we have to find out the number of logs in the top row.
Let the number of rows be n.
So there are (n) terms in the series.
And it is given that the total logs = 200.
So the sum of the series is 200.
Now as we observe the series carefully it will follow the pattern of arithmetic progression series.
With first term (a) = 20, common difference (d) = (19 – 20) = (18 – 19) = -1
And, ${S_n} = 200$, where ${S_n}$ is the sum of the series.
Now as we know that sum of an A.P is given as,
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Now substitute the values we have,
$ \Rightarrow 200 = \dfrac{n}{2}\left( {2\left( {20} \right) + \left( {n - 1} \right)\left( { - 1} \right)} \right)$
Now simplify it we have,
$ \Rightarrow 400 = n\left( {40 - n + 1} \right)$
$ \Rightarrow 400 = - {n^2} + 41n$
$ \Rightarrow {n^2} - 41n + 400 = 0$
Now factorize this we have,
$ \Rightarrow {n^2} - 25n - 16n + 400 = 0$
$ \Rightarrow n\left( {n - 25} \right) - 16\left( {n - 25} \right) = 0$
$ \Rightarrow \left( {n - 25} \right)\left( {n - 16} \right) = 0$
$ \Rightarrow n = 16,25$
So there are two values of n such that the sum of the logs is 200.
Now the last term of the A.P series is calculated as.
$ \Rightarrow {a_n} = a + \left( {n - 1} \right)d$
Now substitute, n = 16 we have,
$ \Rightarrow {a_n} = 20 + \left( {16 - 1} \right)\left( { - 1} \right) = 20 - 15 = 5$
Now when, n = 25 we have,
$ \Rightarrow {a_n} = 20 + \left( {25 - 1} \right)\left( { - 1} \right) = 20 - 24 = - 4$
So when we take a higher value of n there are negative logs in the series which is not possible, as the logs present in the physical world.
So the number of logs in the top row is 5.
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the all the formulas of the A.P which is the basis of this question, so first find out the number of rows in the stack as above then using the last term formula of an A.P calculate the number of logs in the top row.