A ladder has rungs 25 cm apart. 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 $2\dfrac{1}{2}$ m apart, what is the length of the wood required for the rungs?

Given:

Distance between the consecutive rungs = 25 cm

Distance between the top and bottom rungs = $2\dfrac{1}{2}$ m = 2.5 × 100 = 250 cm.

Total number of rungs = $\dfrac{\text{Total length of the ladder}}{\text{Distance between the consecutive rungs}}$ + 1

∴ Total number of rungs = $\dfrac{250}{25} + 1$ = 11.

From the given figure, we can observe that the lengths of the rungs decrease uniformly, hence we can conclude that they will be in an AP

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First-term, a = 45 [length of the lowest rung is 45 cm]

Last term, l = 25 [length of the topmost rung is 25 cm]

Number of terms, n = 11 [Total number of rungs is calculated as 11]

By formula,

Sum of n terms = S_n = $\dfrac{n}{2}$ [First term + Last term]

Substituting values we get :

$S_{11} = \dfrac{11}{2}[45 + 25] \\[1em] = \dfrac{11}{2} \times 70 \\[1em] = 11 \times 35 \\[1em] = 385.$

Hence, the length of the wood required for the rungs is 385 cm.