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?

Given,

20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on.

Let n rows be required to stack 200 logs.

Total logs = 20 + 19 + 18 + ……… upto n terms.

In above sequence, first term (a) = 20 and common difference (d) = 19 - 20 = -1.

By formula,

Sum upto n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow 200 = \dfrac{n}{2}[2 \times 20 + (n - 1) \times (-1)] \\[1em] \Rightarrow 200 = \dfrac{n}{2}[40 - n + 1] \\[1em] \Rightarrow 400 = n[41 - n] \\[1em] \Rightarrow 41n - n^2 = 400 \\[1em] \Rightarrow n^2 - 41n + 400 = 0 \\[1em] \Rightarrow n^2 - 16n - 25n + 400 = 0 \\[1em] \Rightarrow n(n - 16) - 25(n - 16) = 0 \\[1em] \Rightarrow (n - 25)(n - 16) = 0 \\[1em] \Rightarrow n - 25 = 0 \text{ or } n - 16 = 0 \\[1em] \Rightarrow n = 25 \text{ or } n = 16.$

By formula,

a_n = a + (n - 1)d

Substituting values we get :

a₂₅ = 20 + (25 - 1) × (-1)

= 20 + (-24)

= 20 - 24

= -4.

a₁₆ = 20 + (16 - 1) × (-1)

= 20 + (-15)

= 20 - 15

= 5.

Since, logs cannot be negative in a row.

∴ No. of rows cannot be 25.

∴ No. of rows = 16.

Hence, no. of rows = 16 and 5 logs are placed in the top row.