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?
Step-by-Step Solution
Key Concept: $a = 20, d = -1, S_n = 200$. $\\dfrac{n}{2}[2(20) + (n-1)(-1)] = 200 \Rightarrow n(41 - n) = 400 \Rightarrow n^2 - 41n + 400 = 0 \Rightarrow (n - 16)(n - 25) = 0$. For $n = 25$, $a_{25} = 20 + 24(-1) = -4$ (impossible). For $n = 16$, $a_{16} = 20 + 15(-1) = 5$ logs. So $16$ rows and $5$ logs in top row.
$S_n = \dfrac{n}{2}[40 + (n-1)(-1)] = 200 \Rightarrow n(41 - n) = 400 \Rightarrow n^2 - 41n + 400 = 0$. [1.5 Marks]
$(n - 16)(n - 25) = 0 \Rightarrow n = 16$ or $n = 25$. [1.5 Marks]
If $n = 25 \Rightarrow a_{25} = 20 + 24(-1) = -4$ (rejected). If $n = 16 \Rightarrow a_{16} = 20 + 15(-1) = 5$. Thus there are $16$ rows and $5$ logs in the top row. [2.0 Marks]
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🎯 Official CBSE Marking Scheme:
Forming quadratic $n^2 - 41n + 400 = 0$: 1.5 Marks
Solving roots $n = 16, 25$: 1.5 Marks
Rejecting $n = 25$ and concluding $16$ rows with $5$ logs in top row: 2.0 Marks
Correct Answer: