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Arithmetic Progressions
EXERCISE 5.3
CBSE_NCERT_TEXTBOOK
Grade 10
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 (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row? Fig. 5.5
Step-by-Step Solution
Key Concept: The numbers of logs in successive rows form an arithmetic progression (AP) with first term $a=20$, common difference $d=-1$. Use the sum formula for an AP: $S_n = \frac{n}{2}[2a+(n-1)d] = \frac{n}{2}(a+l)$, where $l$ is the last term. Set $S_n = 200$ and solve for $n$.
1. Identify the AP:
\[ a = 20,\quad d = -1 \]
The $n^{th}$ term (top row) is \[ l = a + (n-1)d = 20-(n-1). \]
2. Write the sum of the first $n$ terms (total logs):
\[ S_n = \frac{n}{2}(a + l) = \frac{n}{2}\bigl[20 + (20-(n-1))\bigr] = \frac{n}{2}(41 - n). \]
3. Equate the sum to the given total of logs (200):
\[ \frac{n}{2}(41 - n) = 200 \]
\[ n(41 - n) = 400 \]
\[ -n^{2} + 41n - 400 = 0 \]
\[ n^{2} - 41n + 400 = 0. \]
4. Solve the quadratic equation:
Discriminant $\Delta = 41^{2} - 4\times400 = 1681 - 1600 = 81$.
\[ n = \frac{41 \pm \sqrt{81}}{2} = \frac{41 \pm 9}{2}. \]
Hence $n = \frac{50}{2}=25$ or $n = \frac{32}{2}=16$.
5. Choose the feasible value of $n$.
Since the bottom row has 20 logs, the maximum possible rows are 20 (when the top row would have 1 log). $n=25$ exceeds this limit, so discard it.
Therefore, $n = 16$ rows.
6. Find the number of logs in the top row:
\[ l = 20 - (n-1) = 20 - 15 = 5. \]
7. Verify:
\[ S_{16} = \frac{16}{2}(20+5) = 8 \times 25 = 200 \] (matches the given total).
Thus, the logs occupy 16 rows, and the topmost row contains 5 logs.
\[ a = 20,\quad d = -1 \]
The $n^{th}$ term (top row) is \[ l = a + (n-1)d = 20-(n-1). \]
2. Write the sum of the first $n$ terms (total logs):
\[ S_n = \frac{n}{2}(a + l) = \frac{n}{2}\bigl[20 + (20-(n-1))\bigr] = \frac{n}{2}(41 - n). \]
3. Equate the sum to the given total of logs (200):
\[ \frac{n}{2}(41 - n) = 200 \]
\[ n(41 - n) = 400 \]
\[ -n^{2} + 41n - 400 = 0 \]
\[ n^{2} - 41n + 400 = 0. \]
4. Solve the quadratic equation:
Discriminant $\Delta = 41^{2} - 4\times400 = 1681 - 1600 = 81$.
\[ n = \frac{41 \pm \sqrt{81}}{2} = \frac{41 \pm 9}{2}. \]
Hence $n = \frac{50}{2}=25$ or $n = \frac{32}{2}=16$.
5. Choose the feasible value of $n$.
Since the bottom row has 20 logs, the maximum possible rows are 20 (when the top row would have 1 log). $n=25$ exceeds this limit, so discard it.
Therefore, $n = 16$ rows.
6. Find the number of logs in the top row:
\[ l = 20 - (n-1) = 20 - 15 = 5. \]
7. Verify:
\[ S_{16} = \frac{16}{2}(20+5) = 8 \times 25 = 200 \] (matches the given total).
Thus, the logs occupy 16 rows, and the topmost row contains 5 logs.
Correct Answer: Number of rows = 16; logs in the top row = 5
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