Question

200 logs are stacked in such a way that there are 20 logs in the bottom row, 19 in the next row, 18 in the next row, and so on. In how many rows 200 logs are placed and how many logs are there in the top row?

Answer 1

From the bottom, the number of logs in adjacent rows are 20, 19, 18,...
Since the logs decreases uniformly by 1 in each row, the number of logs in the rows form an AP.
Here, a = 20, d = 19 - 20 = -1
Let there be n rows such that S_n = 200.
The sum of n terms of an AP is given by
$$\begin{gathered} S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ \mathrm{Here},S_n=200 \\ \therefore S_n=200 \\ \Rightarrow \left(\frac{n}{2}\right)\times \left[2\times 20+\left(n-1\right)\times \left(-1\right)\right]=200 \\ \Rightarrow \left(\frac{n}{2}\right)\times [40-n+1]=200 \\ \Rightarrow n\times [41-n]=400 \\ \Rightarrow n^2-41n+400=0 \\ \Rightarrow n^2-25n-16n+400=0 \\ \Rightarrow n(n-25)-16(n-25)=0 \\ \Rightarrow n=25orn=16 \end{gathered}$$

i.e., the number of rows is either 16 or 25.
For n = 25, we have:
T₂₅ = a + 24 d = 20 + 24 ⨯ (-1) = -4, which is not possible.
Therefore, n = 25 is not acceptable.

For n = 16, we have:
T₁₆ = a +15d = 20 + 15 ⨯ (-1)​ = 5
Thus, there are 16 rows and 5 logs are placed in the top row.