NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 19
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 19
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?
Given:
Total number of logs = 200
Number of logs in the bottom row = 20
Number of logs in each subsequent row decreases by 1.
To Find:
Number of rows and number of logs in the top row.
Formula:
The sum of an arithmetic series is given by Sn = n2(2a + (n-1)d). where
n is the number of rows,
a is the number of logs in the top row, and
d is the common difference.
Solution:
In this case,
first term a = 20,
common difference d = -1, and
the total number of logs is 200.
Substituting the values into the formula: n2(2a + (n-1)d).
200 = n2 (2 (20) + (n-1)(-1)).
⇒ 400 = n(40 - n + 1)
⇒ 400 = 41n - n2
⇒ n2 - 41n + 400 = 0
Solve the quadratic equation n2 - 41n + 400 = 0 using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here, a = 1, b = -41, c = 400
Now substitute values:
\( n = \frac{-(-41) \pm \sqrt{(-41)^2 - 4 \times 1 \times 400}}{2 \times 1} \)
\( = \frac{41 \pm \sqrt{1681 - 1600}}{2} \)
\( = \frac{41 \pm \sqrt{81}}{2} \)
\( = \frac{41 \pm 9}{2} \)
So we get two solutions:
\( n = \frac{41 + 9}{2} = \frac{50}{2} = \mathbf{25} \)
\( n = \frac{41 - 9}{2} = \frac{32}{2} = \mathbf{16} \)
Therefore n = 16 or n = 25.
Since there are only 20 logs in the bottom row, n = 25 is not possible.
Therefore, n = 16.
The number of logs in the top row (a) can be calculated as:
a = 20 + (16 - 1)(-1)
a = 20 - 15
a = 5
Result:
The logs are placed in 16 rows, and the top row has 5 logs.
Next question solution:
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 20
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