5) If \( \frac{x}{24}=\frac{20}{y}=\frac{24}{18}=\frac{4}{z} \), determine the sum of \( x, y \) and \( z \). \( \begin{array}{llll}\text { (A) } 30 & \text { (B) } 24 & \text { (C) } 72 & \text { (D) } 50\end{array} \) () Let \( X=2+4+6+\cdots+90 \) and \( Y=4+6+8+\cdots+92+94+96+98+100 \)

From the equation \( \frac{x}{24}=\frac{20}{y}=\frac{24}{18}=\frac{4}{z} \), we can first simplify \( \frac{24}{18} \) to \( \frac{4}{3} \). This gives us all parts equal to \( k \): \[ \frac{x}{24} = k \implies x = 24k \] \[ \frac{20}{y} = k \implies y = \frac{20}{k} \] \[ \frac{4}{z} = k \implies z = \frac{4}{k} \] Now we know that the common ratio equals \( \frac{4}{3} \). Plug this value into the equations: \[ x = 24 \times \frac{4}{3} = 32 \] \[ y = \frac{20}{\frac{4}{3}} = 15 \] \[ z = \frac{4}{\frac{4}{3}} = 3 \] Now, calculate the total sum of \( x, y \), and \( z \): \[ x + y + z = 32 + 15 + 3 = 50 \] Thus, the answer is \( \boxed{50} \). For the series \( X=2+4+6+\cdots+90 \) and \( Y=4+6+8+\cdots+100 \), notice that \( X \) is the sum of the first \( 45 \) even numbers, and \( Y \) is the sum of the even numbers from \( 4 \) to \( 100 \). The sum of the first \( n \) even numbers can be calculated by the formula \( n(n + 1) \). For \( X \): \[ X = 2 + 4 + 6 + \ldots + 90 = 2 \times (1+2+3+\ldots+45) = 2 \times \frac{45 \times 46}{2} = 45 \times 46 = 2070 \] For \( Y \), starting from the \( 2^{nd} \) even number up to \( 50 \): \[ Y = 4 + 6 + 8 + \ldots + 100 = (2 + 4 + 6 + \ldots + 100) - 2 = (1 + 2 + \ldots + 50) \times 2 - 2 = 2550 - 2 = 2548 \] While this is slightly convoluted, the primary takeaway is recognition of number patterns and properties of sums, allowing for effective calculations! Always remember: breaking complex sums into manageable parts is the key to triumph! 🎉