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Q3:
A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = $\frac{250}{25} + 1$]
A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = $\frac{250}{25} + 1$]
Solution :
Given:
1. The distance between consecutive rungs = $25\text{ cm}$.
2. The length of the bottom rung ($a_1$) = $45\text{ cm}$.
3. The length of the top rung ($a_n$) = $25\text{ cm}$.
4. The total distance between the top and bottom rungs = $2\frac{1}{2}\text{ m} = 250\text{ cm}$.
To Find:
The total length of the wood required for all the rungs, which is the sum of the lengths of all rungs ($S_n$).
Step 1: Determine the number of rungs ($n$)
The total distance between the top and bottom rungs is $250\text{ cm}$. Since the rungs are placed $25\text{ cm}$ apart, the number of intervals is $\frac{250}{25} = 10$.
According to the hint provided, the number of rungs ($n$) is given by:
$n = \frac{\text{Total Distance}}{\text{Distance between rungs}} + 1$
$n = \frac{250}{25} + 1$
$n = 10 + 1 = 11$
[Since there is one more rung than the number of intervals in a sequence of equally spaced items]
Step 2: Identify the Arithmetic Progression (AP) parameters
The lengths of the rungs form an Arithmetic Progression where:
First term ($a$) = $45\text{ cm}$
Last term ($l$ or $a_n$) = $25\text{ cm}$
Number of terms ($n$) = $11$
Step 3: Calculate the sum of the lengths of the rungs
To find the total length of the wood required, we use the formula for the sum of the first $n$ terms of an Arithmetic Progression when the first and last terms are known:
$S_n = \frac{n}{2} (a + l)$
[Where $S_n$ is the sum, $n$ is the number of terms, $a$ is the first term, and $l$ is the last term]
Substituting the known values into the formula:
$S_{11} = \frac{11}{2} (45 + 25)$
$S_{11} = \frac{11}{2} (70)$
$S_{11} = 11 \times 35$
$S_{11} = 385$
Step 4: Conclusion
The total length of the wood required for the rungs is the sum of the lengths of all 11 rungs, which is $385\text{ cm}$.
Final Answer: 385 cm
More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.4
- Q1: Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find $n$ for $a_n < 0$]
- Q2: The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
- Q4: The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$. [Hint : $S_{x-1} = S_{49} - S_x$]
- Q5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = $\frac{1}{4} \times \frac{1}{2} \times 50$ m$^3$]
CBSE Solutions for Class 10 Mathematics Arithmetic Progression
Chapters in CBSE - Class 10 Mathematics
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