# Q : 19      $\small 200$  logs are stacked in the following manner: $\small 20$ logs in the bottom row, $\small 19$  in the next row, $\small 18$ in the row next to it and so on (see Fig. $\small 5.5$). In how many rows are the $\small 200$ logs placed and how many logs are in the top row?

G Gautam harsolia

As, the rows are going up, the no of logs are decreasing,
We can clearly see that 20, 19, 18, ..., is an AP .
and here  $a = 20 \ and \ d = -1$
Let suppose 200 logs are arranged in 'n' rows,
Then,
$S_n = \frac{n}{2}\left \{ 2\times 20 +(n-1)(-1) \right \}$
$200 = \frac{n}{2}\left \{ 41-n \right \}$
$\Rightarrow n^2-41n +400 = 0$
$\Rightarrow n^2-16n-25n +400 = 0$
$\Rightarrow (n-16)(n-25) = 0$
$\Rightarrow n = 16 \ \ and \ \ n = 25$
Now,
case (i) n = 25
$a_{25} =a+24d = 20+24\times (-1)= 20-24=-4$
But number of rows can not be in negative numbers
Therefore, we will reject the value n = 25

case (ii) n = 16

$a_{16} =a+15d = 20+15\times (-1)= 20-15=5$
Therefore, the number of rows in which 200 logs are arranged is equal to 5

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