NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.4 (Optional)

Question 3

Step 1: Convert the distance between top and bottom rungs into cm.

Distance between top and bottom rungs is \(2\dfrac{1}{2}\) m:

\[2\dfrac{1}{2} = 2.5\,\text{m} = 2.5 \times 100 = 250\,\text{cm}\]

Step 2: Find the number of rungs.

The rungs are 25 cm apart. So the number of intervals between top and bottom rungs is:

\[\frac{250}{25} = 10\]

Number of rungs = number of intervals + 1:

\[\text{Number of rungs} = 10 + 1 = 11\]

Step 3: Observe that rung lengths form an AP.

The lengths decrease uniformly from 45 cm at the bottom to 25 cm at the top. So the rung lengths form an arithmetic progression:

\[45, \ldots, 25\]

First term (bottom rung): \(a = 45\,\text{cm}\)

Last term (top rung): \(l = 25\,\text{cm}\)

Number of terms (rungs): \(n = 11\)

Step 4: Use the sum formula for an AP.

Sum of \(n\) terms of an AP:

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

Here,

\[S_{11} = \frac{11}{2}(45 + 25) = \frac{11}{2} \times 70\]

\[S_{11} = 11 \times 35 = 385\,\text{cm}\]

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