Question

$200$ 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 (see figure) In how many rows are the $200$ logs placed and how many logs are in the top row?

Answer 1

Total no. of logs ($S_n)=200$

No. of logs in the first row = $20$

No. of logs in the second row = $19$

No. of logs in the second row = $18$

Thus, $20,19,\mathrm{18.....}$ forms an AP with,

First term $\left(a\right)=20$ and common difference $\left(d\right)=19-20=-1$

Now using the formula, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$, we get,

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

$\Rightarrow 200=\frac{n}{2}[40-n+1]$

$\Rightarrow 200\times 2=41n-n^2$

$\Rightarrow n^2-41n+400=0$

$\Rightarrow n^2-16n-25n+400=0$

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

Hence, $n=25,16$

If, $n=16$

Then, top row $\left(a_{16}\right)=a+(n-1)d=20+15(-1)=5$

If, $n=25$

Then, top row $\left(a_{25}\right)=a+(n-1)d=20+24(-1)=-4$

Since, a row cannot be negative, therefore ,$n=16$

Hence, there are $16$ rows with $5$ logs at the top row.