A smaller terrace at a football …

Volume of concrete required to build the first step, second step, third step, ..... (in m²) are
{tex}\frac{1}{4} \times \frac{1}{2} \times 50,\left( {2 \times \frac{1}{4}} \right) \times \frac{1}{2} \times 50,\left( {3 \times \frac{1}{4}} \right) \times \frac{1}{2} \times 50{/tex}
{tex} \Rightarrow \frac{{50}}{8},2 \times \frac{{50}}{8},3 \times \frac{{50}}{8},.....{/tex}
{tex}\therefore {/tex} Total volume of concrete required = {tex}\frac{{50}}{8} + 2 \times \frac{{50}}{8} + 3 \times \frac{{50}}{8} + ....{/tex}
{tex} = \frac{{50}}{8}\left[ {1 + 2 + 3 + .......} \right]{/tex}
S_n = {tex}\frac{{n}}{2}{/tex} [(2a + (n - 1)d]
S₁₅  {tex} = \frac{{50}}{8} \times \frac{{15}}{2}\left[ {2 \times 1 + (15 - 1) \times 1} \right]\left[ {\because n = 15} \right]{/tex}
{tex} = \frac{{50}}{8} \times \frac{{15}}{2} \times 16{/tex}
= 750 m³