Question

The number of shots in a triangular pile is greater by 150 than half the number of shots in a square pile, the number of layers in each being the same; find the number of shots in the lowest layer of the triangular pile.

Answer 1

Hint: First we will first assume that \[n\] represents the number of layers and then use the number of shot in triangle pile is \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}\] and the number of shot in triangle pile is \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]. Then we will subtract the half of the number of shots in square pile from the number of shots in triangular pile to find the required value.

Complete step by step answer:

We are given that the number of shots in a triangular pile is greater by 150 than half the number of shots in a square pile, the number of layers in each being the same.

Let us assume that \[n\] represents the number of layers.

We know that the number of shot in triangle pile is \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}\] and the number of shot in square pile is \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\].

Subtracting the half of the number of shots in square pile from the number of shots in triangular pile, we get

\[
   \Rightarrow \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6} - \dfrac{1}{2} \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} = 150 \\
   \Rightarrow \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6} - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{12}} = 150 \\
   \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{n + 2}}{3} - \dfrac{{2n + 1}}{6}} \right) = 150 \\
   \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{2n + 4 - 2n - 1}}{6}} \right) = 150 \\
   \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{3}{6}} \right) = 150 \\
   \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{1}{2}} \right) = 150 \\
 \]

Multiplying the above equation by 2 on both sides, we get
\[ \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2} = 300\]

Hence, the final answer is 300.

Note: In solving this type of question, the key concept is to know that the formula to find number of shots in a square pile is \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}\] and the number of shots in a triangular pile is \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}\]. Then this problem is really easy to solve. One writes the formula of triangular pile for both, which is wrong.