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. In how many rows are the $200$logs placed and how many logs are in the top row.

Answer 1

Hint: Given that 1st row have $20$logs, 2nd row have $19$logs , 3rd row have $18$and so on. Hence the series is $20$, $19$,$18$ and so on. Since, the difference is the same it is in an arithmetic progression. Here, we will use the formula for the sum of n terms in arithmetic progression.

Formula Used:
${S_n} = \dfrac{n}{2}\left[ {a + \left( {n - 1} \right)d} \right]$
Where, $a$ is the first term of the series, $d$is the difference. Our aim is to find the value of $n$.

Complete step by step solution
Given:

Number of logs in the 1st row is $20$.

Number of logs in the 2nd row is $19$.

Number of logs in the 3rd row is $18$.

The series formed is given as,

$20$, $19$, $18$,....

Since the difference is the same, then we can say that it is an arithmetic progression.

We need to find how many rows can $200$ logs can be placed. We can use formula for arithmetic progressions, that is,
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$

Where, ${S_n}$ is the total sum which is equal to $200$, $a$ is the first term which is equal to $20$, $d$ is the difference between the terms and it is equal to $ - 1$.

On substituting the values of ${S_n}$, $a$ and $d$ in the above expression, we get,
$\begin{array}{l}
200 = \dfrac{n}{2}\left( {2 \times 20 + \left( {n - 1} \right) \times - 1} \right)\\
200 = \dfrac{n}{2}\left( {40 - n + 1} \right)\\
200 = \dfrac{n}{2}\left( {41 - n} \right)
\end{array}$

Then on rearranging the terms we get,
$\begin{array}{c}
n\left( {41 - n} \right) = 400\\
{n^2} - 41n + 400 = 0
\end{array}$

We will factorize the above equation, we get,
$\begin{array}{c}
{n^2} - 16n - 25n + 400 = 0\\
n\left( {n - 16} \right) - 25\left( {n - 16} \right) = 0
\end{array}$

Hence, the value of $n$ is $16$ or $25$.

Therefore, the number of rows is $16$ but not $25$ because the number of logs at $25$ will be $ - 4$which is not possible.


Note: The one of the important points in this problem is that there are $20$ logs in 1st row and $19$ rows in 2nd row and $18$ logs in 3rd row and so on. There will be $5$logs in $16$ row and there will be $ - 4$ logs in $25$ row. Since, $ - 4$ logs are not possible. So, $25$ number of rows is not a possible solution.