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

## Question

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.

## 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.

**Reasoning:**

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.

**Steps:**

From the figure, it can be observed that

Height of 1^{st} 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.

\[6.25\,\rm{m^3},12.50\,\rm{m^3},\,18.75\,\rm{m^3},.......\]

- 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 \(750\;\rm{m^3}\).