Ex.5.4 Q5 Arithmetic progressions Solutions - NCERT Maths Class 10

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A small terrace at a football ground comprises of \(15\) steps each of which is \(50\, \rm{m}\) long and built of solid concrete. Each step has a rise of \(\begin{align}\frac{1}{4}\,\rm{m} \end{align}\) and a tread of \(\begin{align}\frac{1}{2}\,\rm{m} \end{align}\) (See figure) calculate the total volume of concrete required to build the terrace.

 Video Solution
Arithmetic Progressions
Ex 5.4 | Question 5

Text Solution

What is Known?

\(15\) steps each of which is \(50 \,\rm{m}\) long and each step has a rise of \(\begin{align} \frac{1}{4}\,\rm{m} \end{align}\) and a tread of \(\begin{align} \frac{1}{2}\,\rm{m} \end{align}\)

What is Unknown?

Total volume of concrete required to build the terrace.


Sum of the first \(n\) terms of an AP is given by \(\begin{align} {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \end{align}\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.


From the figure, it can be observed that

Height of 1st step is \(\begin{align}\frac{1}{4} \,\rm{m} \end{align}\)

Height of 2nd step is \(\begin{align} \left( {\frac{1}{4} + \frac{1}{4}} \right) \,\rm{m} = \frac{1}{2} \,\rm{m} \end{align}\)

Height of 3rd step is \(\begin{align} \left( {\frac{1}{2} + \frac{1}{4}} \right)\,\rm{m} = \frac{3}{4} \,\rm{m} \end{align}\)

Therefore, height of the each step is increasing by \(\begin{align} \frac{1}{4} \,\rm{m} \end{align}\)  length \(50 \,\rm{m}\) and width (tread) \(\begin{align} \frac{1}{2} \,\rm{m} \end{align}\) remain the same for each of the steps.

Volume of Step can be considered as \( \text{Volume of Cuboid}= Length \times Breadth \times Height\)

Volume of concrete in 1st step \(\begin{align} = 50m \times \frac{1}{2} \,\rm{m} \times \frac{1}{4} \,\rm{m} = 6.25\,{m^3} \end{align}\)

Volume of concrete in 2nd step \(\begin{align} = 50 \,\rm{m} \times \frac{1}{2} \,\rm{m} \times \frac{1}{2} \,\rm{m} = 12.50\,{m^3} \end{align}\)

Volume of concrete in 3rd step \(\begin{align} = 50 \,\rm{m} \times \frac{1}{2} \,\rm{m} \times \frac{3}{4} \,\rm{m} = 18.75\,{m^3} \end{align}\)

It can be observed that the volumes of concrete in these steps are in an A.P.


  • First term, \(a = 6.25\)
  • Common difference, \(d = 6.25\)
  • Number of steps, \(n = 15\)

Sum of n terms, \(\begin{align} {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \end{align}\)

\[\begin{align}{S_{15}} &= \frac{{15}}{2}\left[ {2 \times 6.25 + \left( {15 - 1} \right) \times 6.25} \right]\\&= \frac{{15}}{2}\left[ {12.50 + 14 \times 6.25} \right]\\ &= \frac{{15}}{2}\left[ {12.50 + 87.50} \right]\\ &= \frac{{15}}{2} \times 100\\& = 750\end{align}\]

Therefore, Volume of concrete required to build the terrace is \(750\;\rm{m^3}\).

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