Question

I:=2; a:=6; while i<5 do begin a:=a+i; i:=i+1; end

Answer 1

I = 2.
A = 6.
1. A = 6 + 2 = 8.
I = 2 + 1 = 3.

2. A = 8 + 3 = 11.
I = 3 + 1 = 4.

3. A = 11 + 4 = 15.
I = 4 + 1 = 5.

Ответ: A = 15, I = 5.

Answer 2

It seems like you are trying to write a simple loop in a programming language. Based on the syntax you provided, it appears to be Pascal or a similar language. The code you've written is a loop that will execute as long as the variable i is less than 5. In each iteration, the value of a will be updated by adding the current value of i, and then i will be incremented by 1.

Let's go through the code step-by-step:

1. I:=2;: This line initializes the variable i with the value 2.

2. a:=6;: This line initializes the variable a with the value 6.

3. while i<5 do: This is the start of the loop. As long as the value of i is less than 5, the loop will continue to execute.

4. begin: This is the beginning of the loop block, where the loop statements are enclosed.

5. a:=a+i;: In each iteration of the loop, the value of a will be updated by adding the current value of i to it.

6. i:=i+1;: After updating a, the value of i will be incremented by 1.

7. end: This marks the end of the loop block.

After going through the loop, the value of a will be the sum of the initial value of a (6) and the values of i in each iteration (2+3+4 = 9). Since the loop condition is i < 5, the loop will execute three times (i=2, i=3, i=4).

So, after the loop completes, the final values will be:

i = 5 (as it becomes 5 after the loop) a = 15 (6 + 2 + 3 + 4)