$228$ logs are to be stacked in a store in the following manner: $30$ logs in the bottom, $28$ logs in the next row, then $26$ and so on, in how many rows can these $228$ logs be stacked? How many logs are there in the last row?

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Solution

Step 1. Given:
We have,
Number of logs in the bottom row $= 30$
Number of logs in $2^{n d}$ row from the bottom row $= 28$
Number of logs in the $3^{r d}$ row from the bottom row $= 26$
Now, the number of logs in the successive rows from bottom are $30, 28, 26, \dots .$
Clearly, this sequence of numbers forms an AP.
For this AP, first term, $a = 30$ and common difference, $d = 28 - 20 = - 2$
Let the top row be $n^{t h}$ row.
Step 2. Find the value of $n$ :
Formula:
Sum of first $n$ terms of AP is $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , where $a$ is the first term and $d$ is the common ratio.
According to the question, we get $S_{n} = 228$ .
$\begin{matrix} \Rightarrow & \frac{n}{2} [2 a + (n - 1) d] = 228 \\ \Rightarrow & \frac{n}{2} [2 \times 30 + (n - 1) \times (- 2)] = 228 \\ \Rightarrow & n [30 + (n - 1) \times (- 1)] = 228 \\ \Rightarrow & n [30 - n + 1] = 228 \\ \Rightarrow & 30 n - n^{2} + n = 228 \\ \Rightarrow & n^{2} - 31 n + 228 = 0 \end{matrix}$
On solving the quadratic equation, we get
$\begin{matrix} \Rightarrow & n^{2} - 19 n - 12 n + 228 = 0 \\ \Rightarrow & (n - 19) (n - 12) = 0 \\ \Rightarrow & n = 12, 19 \end{matrix}$
Step 3: Find the last row value $l$ :
Formula:
The $n^{th}$ term of AP is $l = a + (n - 1) d$ , where $a$ is the first term and $d$ is the common ratio and $l$ is the last term.
For $n = 19$ :
$\begin{array}{rcl} l & = & 30 + (19 - 1) \times (- 2) \\ = & 30 - 36 \\ = & - 6 \end{array}$
$∵$ the number of logs cannot be negative, the number of rows should be 12.
For $n = 12$ :
$\begin{array}{rcl} l & = & 30 + (12 - 1) \times (- 2) \\ = & 30 - 22 \\ = & 8 \end{array}$
Hence, the number of rows is $12$ and the last row is $8 .$

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