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Arithmetic Progressions
CH05 Question Bank
CBSE_CH05_QUESTION_BANK
Grade 10

Question:

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 bottom rungs are $2\dfrac12$ m apart, find the length of the wood required for the rungs, and also the total number of rungs.

Step-by-Step Solution

Key Concept: Recognise that the rung lengths form an AP; first find the number of rungs using the total vertical span, then use the sum formula to find total wood needed.
Total vertical distance $=2\dfrac12$ m $=250$ cm, and consecutive rungs are $25$ cm apart, so the number of gaps $=\dfrac{250}{25}=10$, meaning there are $11$ rungs in total. [1.5 Marks]

The rung lengths form an AP with first term $a=45$ cm, last term $l=25$ cm, and $n=11$ terms. [1.0 Mark]

Total wood required $=S_{11}=\dfrac{11}{2}(a+l)=\dfrac{11}{2}(45+25)$. [1.5 Marks]

$=\dfrac{11}{2}\times70=11\times35=385$ cm. So $385$ cm of wood is needed for all $11$ rungs. [1.0 Mark]

Correct Answer:
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