Consider the following code segment, which is intended to store ten consecutive even integers, beginning with 2, in the list evenlist. Assume that evenList is initially empty. iarrow 1 REPEAT 10 TIMES MISSING CODE> Which of the following can be used to replace <MISSING CODE> SO that the code segment works as intended? A APPEND(evenList, 1) iarrow i+2 B iarrow i+2 APPEND(evenList, i) C (c) APPEND(even List, 2ast i iarrow i+1 D iarrow i+1 APPEN ocev enLi st, 2ast i

To solve this problem, we need to understand the requirements of the code segment and the options provided.<br /><br />The code segment is intended to store ten consecutive even integers, starting with 2, in the list evenList. The initial value of i is 1, and the code segment is repeated 10 times.<br /><br />Let's analyze the options:<br /><br />A. APPEND(evenList, 1)<br /> i <- i + 2<br /> This option appends the number 1 to the list, which is not an even integer. Therefore, this option is incorrect.<br /><br />B. i <- i + 2<br /> APPEND(evenList, i)<br /> This option increments the value of i by 2 and then appends the value of i to the list. However, this will not produce the correct sequence of even integers. Therefore, this option is incorrect.<br /><br />C. APPEND(evenList, 2*i)<br /> i <- i + 1<br /> This option appends the value of 2*i to the list and then increments the value of i by 1. This will produce the correct sequence of even integers. Therefore, this option is correct.<br /><br />D. i <- i + 1<br /> APPEND(evenList, 2*i)<br /> This option increments the value of i by 1 and then appends the value of 2*i to the list. This will also produce the correct sequence of even integers. Therefore, this option is correct.<br /><br />Therefore, the correct answer is option C.