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Arithmetic Progressions | Arithmetic Progressions - Exercise 5.4

Question 3

A ladder has rungs 25 cm apart. (See figure). 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\ \frac{1}{2} m apart, what is the length of the wood required for the rungs?

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Given,

Distance between the rungs of the ladder is 25cm.

Distance between the top rung and bottom rung of the ladder is = 2\frac{1}{2}\mathrm{m}=2\frac{1}{2}\times100\mathrm{cm}=5/2\times100\mathrm{cm}=250cm

Therefore, total number of rungs = 250/25 + 1 = 11

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

So,

First term, a = 45

Last term, l = 25

Number of terms, n = 11

Now, as we know, sum of nth terms is equal to,

\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+1) \\ \therefore \mathrm{S}_{10} \frac{11}{2}(45+25)=\frac{11}{2}(70)=385 \mathrm{~cm}

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

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Question 3

A ladder has rungs 25 cm apart. (See figure). 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\ \frac{1}{2} m apart, what is the length of the wood required for the rungs?

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given,

Distance between the rungs of the ladder is 25cm.

Distance between the top rung and bottom rung of the ladder is = 2\frac{1}{2}\mathrm{m}=2\frac{1}{2}\times100\mathrm{cm}=5/2\times100\mathrm{cm}=250cm

Therefore, total number of rungs = 250/25 + 1 = 11

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

So,

First term, a = 45

Last term, l = 25

Number of terms, n = 11

Now, as we know, sum of nth terms is equal to,

\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+1) \\ \therefore \mathrm{S}_{10} \frac{11}{2}(45+25)=\frac{11}{2}(70)=385 \mathrm{~cm}

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

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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