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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 = $\frac{250}{25} + 1$]

Solution :

Given:

1. The distance between consecutive rungs is $d_{gap} = 25$ cm.

2. The length of the bottom rung ($a_1$) = $45$ cm.

3. The length of the top rung ($a_n$) = $25$ cm.

4. The total distance between the top and bottom rungs = $2\frac{1}{2}$ m = $250$ 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: Determining the total number of rungs ($n$)

The total number of rungs is calculated by dividing the total distance by the gap between rungs and adding 1 (to account for the starting rung).

$n = \frac{\text{Total Distance}}{\text{Gap between rungs}} + 1$

$n = \frac{250}{25} + 1$

$n = 10 + 1 = 11$

[Since there are 11 intervals of 25 cm to cover 250 cm, there must be 11 rungs.]


Step 2: Identifying the Arithmetic Progression (AP) parameters

Let the lengths of the rungs form an Arithmetic Progression where:

First term ($a$) = $45$ cm

Last term ($l$ or $a_n$) = $25$ cm

Number of terms ($n$) = $11$


Step 3: Calculating the total length of wood required

The total length of wood required is the sum of the lengths of all $n$ rungs. We use the formula for the sum of an AP when the first and last terms are known:

$S_n = \frac{n}{2} (a + l)$

[Where $S_n$ is the sum of $n$ terms, $a$ is the first term, and $l$ is the last term.]

Substituting the known values:

$S_{11} = \frac{11}{2} (45 + 25)$

$S_{11} = \frac{11}{2} (70)$

$S_{11} = 11 \times 35$

$S_{11} = 385$


Step 4: Verification of units

Since the lengths were provided in centimeters (cm), the sum is also in centimeters.

Total length = $385$ cm.


Final Answer: The total length of the wood required for the rungs is 385 cm.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.4 (Optional)*


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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