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

We can consider the number of logs in each row from bottom as an AP with 1st term 20 and common difference -1. Then we can write the equation for the sum of n terms and equate it to 200. Then we can solve for n to get the number of rows. We can find the number of logs in the top row by calculating the nth term of the AP.

Complete step by step solution:
We are given that there are 20 logs in the bottom row.
$\Rightarrow {a_1} = 20$
The next row has 19 logs and the row next to it has 18 logs.
$\Rightarrow {a_2} = 19$
$\Rightarrow {a_3} = 18$
Thus, the number of logs in each row forms an AP with 1st term ${a_1} = 20$
To find the common difference, we can subtract a term in the AP with the previous term.
$\Rightarrow d = {a_3} - {a_2}$
On substituting the values we get,
$\Rightarrow d = 18 - 19$
On simplification we get,
$\Rightarrow d = - 1$
Now we have an AP with 1st term $a = 20$ and common difference $d = - 1$
We know that sum of n terms of an AP is given by,
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
We can substitute the values of a and d.
$\Rightarrow {S_n} = \dfrac{n}{2}\left( {2 \times 20 + \left( {n - 1} \right)\left( { - 1} \right)} \right)$
On simplification, we get,
$\Rightarrow {S_n} = \dfrac{n}{2}\left( {41 - n} \right)$
It is given that there are 200 logs to be arranged.
$\Rightarrow 200 = \dfrac{n}{2}\left( {41 - n} \right)$
On multiplying throughout with 2, we get,
$\Rightarrow 400 = n\left( {41 - n} \right)$
We can expand the bracket and rearrange,
$\Rightarrow {n^2} - 41n + 400 = 0$
We can split the middle term,
$\Rightarrow {n^2} - 25n - 16n + 400 = 0$
Now we can take the common factors
$\Rightarrow n\left( {n - 25} \right) - 16\left( {n - 25} \right) = 0$
On taking $\left( {n - 16} \right)$ common we get,
$\Rightarrow \left( {n - 16} \right)\left( {n - 25} \right) = 0$
Hence we have,
$\Rightarrow n = 16,25$
We know that nth term of an AP is given by the equation,
${a_n} = a + \left( {n - 1} \right)d$
When n is 16, 16th term will be,
$\Rightarrow {a_{16}} = 20 + \left( {16 - 1} \right)\left( { - 1} \right)$
$\Rightarrow {a_{16}} = 20 - 15$
$\Rightarrow {a_{16}} = 5$
When n is 25, 25th term will be,
$\Rightarrow {a_{25}} = 20 + \left( {25 - 1} \right)\left( { - 1} \right)$
$\Rightarrow {a_{25}} = 20 - 24$
$\Rightarrow {a_{25}} = - 4$
As the terms of AP represents the number of logs, it cannot be negative.
So, we can reject $n = 25$
So, the number of rows is 16 and the number of logs in the top row is 5.

Note: While calculating the common difference, we must subtract the nth term from the $n + 1$ th term. We may take the reverse order as the difference is negative. The common difference of an AP can be negative if the series is decreasing. While solving for n, we get a quadratic equation and it will have 2 solutions. We must take only the positive value of n as the number of terms cannot be negative. If both values of n are positive, we must check the value of nth term. In this case the terms of the AP cannot be negative as they represent the number of logs in each row.