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# A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $${{1} \over {4}}$$ m and a tread of $${{1} \over {2}}$$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = $${{1} \over {4}} × {{1} \over {2}} × 50 m^3$$.]

As we can see from the given figure, the first step is $${{1} \over {2}}$$ m wide, 2nd step is 1 m wide and 3rd step is $${{3} \over {2}}$$ m wide. Thus we can understand that the width of step by $${{1} \over {2}}$$ m each time when height is $${{1} \over {4}}$$ m. And also, given length of the steps is 50 m all the time. So, the width of steps forms a series AP in such a way that;

$${{1} \over {2}}$$ , 1, $${{3} \over {2}}$$, 2, ……..
Volume of steps = Volume of Cuboid
Now,

Volume of concrete required to build the first step = $${{1} \over {4}} × {{1} \over {2}} ×50 = {{25} \over {4}}$$

Volume of concrete required to build the second step = $${{1} \over {5}} × {{2} \over {2}} × 50 = {{25} \over {2}}$$

Volume of concrete required to build the second step = $${{1} \over {4}} × {{3} \over {2}}×50 = {{75} \over {2}}$$

Now, we can see the volumes of concrete required to build the steps, are in AP series;

$${{25} \over {4}} , {{25} \over {2}} , {{75} \over {2}}$$…..
Thus, applying the AP series concept,

First term, a = $${{25} \over {4}}$$
Common difference, d = $${{25} \over {2}} – {{25} \over {4}} = {{25} \over {4}}$$

As we know, the sum of n terms is;

$$S_n = {{n} \over {2}}[2a+(n-1)d]$$
$$= {{15} \over {2}}(2×({{25} \over {4}} )+({{15} \over {2}} -1){{25} \over {4}})$$

Upon solving, we get,

$$S_n = {{15} \over {2}} (100)$$
$$S_n= 750$$

Hence, the total volume of concrete required to build the terrace is $$750 m^3$$.