200 logs are stacked in the following manner :

Solution:

It can be observed that the numbers of logs in rows are in an A.P.

$20,19,18 \ldots$

For this A.P.,

a = 20

$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=19-20=-1$

Let a total of 200 logs be placed in n rows.

$S_{n}=200$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$200=\frac{n}{2}[2(20)+(n-1)(-1)]$

$400=n(40-n+1)$

$400=n(41-n)$

$400=41 \mathrm{n}-\mathrm{n}^{2}$

$n^{2}-41 n+400=0$

$n^{2}-16 n-25 n+400=0$

$n(n-16)-25(n-16)=0$

$(n-16)(n-25)=0$

Either (n – 16) = 0 or n – 25 = 0

n = 16 or n = 25

$a_{n}=a+(n-1) d$

$a_{16}=20+(16-1)(-1)$

$a_{16}=20-15$

$a_{16}=5$

Similarly,

$a_{25}=20+(25-1)(-1)$

$a_{25}=20-24$

$=-4$

Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which is not possible.

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