Question

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row and 18 in the row next to it and so on as in the figure. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer 1

Hint: Here, we will get an arithmetic series 20, 19, 18, … with common difference -1 . Since ${{S}_{n}}$ is given as 200 therefore, we can find the value of n by the formula, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$. Here, we will get a quadratic equation, which has two roots. From the two values of n find the possible value. Then, calculate the number of logs in the top row by the formula, ${{a}_{n}}=a+\left( n-1 \right)d$.

Complete step-by-step answer:
Here, it is given that 200 logs are stacked in such a way that 20 logs in the bottom row, 19 in the next row, 18 in the next row and so on.
Hence the series is:
20, 19, 18, .….
Since, the common difference is the same, i.e. d = -1, we can say that the above series is in AP with the first term a = 20.
Now, we have to find out how many rows are the 200 logs placed and how many logs in the top row.
So, we have to find the value of n and the last term of the series.
We are given that the sum of the series ${{S}_{n}}=200$
Now, consider the formula for ${{S}_{n}}$.
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
We have ${{S}_{n}}=200,a=20,d=-1$ .
Now, by substituting all these values in the formula we get:
$\begin{aligned} & 200=\dfrac{n}{2}\left[ 2\times 20+(n-1)\times -1 \right] \\\ \Rightarrow & 200=\dfrac{n}{2}\left[ 40-n+1 \right] \\\ \Rightarrow & 200=\dfrac{n}{2}\left[ 41-n \right] \\\ \end{aligned}$
Next, by cross multiplication we get:
$\begin{aligned} & 200\times 2=n\left[ 41-n \right] \\\ \Rightarrow & 400=n\left[ 41-n \right] \\\ \Rightarrow & 400=n\times 41-n\times n \\\ \Rightarrow & 400=41n-{{n}^{2}} \\\ \end{aligned}$
Now, by taking all the terms from right side to left side i.e. 41n becomes -41n and –n2 becomes n2. Hence we get the equation:
$\Rightarrow {{n}^{2}}-41n+400=0$
Therefore, we can say that the above equation is a quadratic equation.
Now, we have to find the two roots of the quadratic equation. For that split the term – 41n as –16n – 25n. Hence, we get the equation:
$\Rightarrow {{n}^{2}}-16n-25n+400=0$
Next, we have to take the common factor n from the first two terms and the common factor -25 from the last two terms outside. We obtain the equation:
$\Rightarrow n(n-16)-25(n-16)=0$
Since, n-16 is the common factor we can take it outside. Hence we get:
$\Rightarrow (n-16)(n-25)=0$
We know that if $a.b=0$ then either a = 0 or b = 0. Hence we obtain:
$\Rightarrow n-16=0$ or $n-25=0$
Now, consider:
$n-16=0$
By taking -16 to the right side it becomes 16. We will get:
$n=16$
Next, consider:
$n-25=0$
By taking -25 to the right side it becomes 25. We will get:
$n=25$
Hence, $n=16$ or $n=25$
To identify the number of logs we have the formula to find the nth term which is given as:
${{a}_{n}}=a+\left( n-1 \right)d$
For n = 16 we have:
$\begin{aligned} & {{a}_{16}}=20+\left( 16-1 \right)\times -1 \\\ \Rightarrow & {{a}_{16}}=20+15\times -1 \\\ \Rightarrow & {{a}_{16}}=20-15 \\\ \Rightarrow & {{a}_{16}}=5 \\\ \end{aligned}$
For n = 25 we have:
$\begin{aligned} & {{a}_{25}}=20+\left( 25-1 \right)\times -1 \\\ \Rightarrow & {{a}_{25}}=20+24\times -1 \\\ \Rightarrow & {{a}_{25}}=20-24 \\\ \Rightarrow & {{a}_{25}}=-4 \\\ \end{aligned}$
Since, -4 is not possible so we cannot take n = 25
Hence, the value of n is 16.
Therefore, we can say that 200 logs are placed in 16 rows and the top row is the ${16}^{th}$ row and it has 5 logs.

Note: Here, the number of logs cannot be a negative value, so you can’t take n as 25. If you consider n = 25 then you will get a wrong answer. So, in case of two values, you should always consider the value which is appropriate for the conditions provided for the solution.