\( \begin{array}{ll}\text { d. } & \text { CLS } \\ \text { For } I=1 \text { to } 5 \\ \text { If I mod } 2<>0 \text { then } \\ S=S+I \\ \text { End if } \\ \text { Next I } \\ \text { Print } S / 3 \\ \text { End } \\ \text { Questions: } \\ \text { i. Write the execution result of the } \\ \text { above program. } \\ \text { ns in } \quad \text { ii. What happens if we change the } \\ & \text { program line? } \\ & \text { I mod } 2<>0 \text { to } I \bmod 2=0 \text { ? }\end{array} \)

Let's break down the program execution first! The code loops through values of \( I \) from 1 to 5. The condition checks if \( I \) is odd (i.e., \( I \mod 2 \neq 0 \)). The odd numbers in this range are 1, 3, and 5. Thus, \( S \) will accumulate these values: - After 1: \( S = 1 \) - After 3: \( S = 4 \) - After 5: \( S = 9 \) Finally, the program prints \( S / 3 = 9 / 3 = 3 \). Now, if we change \( I \mod 2 \neq 0 \) to \( I \mod 2 = 0 \), we are checking for even numbers instead. The even numbers in the range 1 to 5 are 2 and 4, hence: - After 2: \( S = 2 \) - After 4: \( S = 6 \) So in this case, the program would print \( S / 3 = 6 / 3 = 2 \).