Answer :
Sure! Let's walk through the program step by step and see the detailed solution.
1. We initialize two variables [tex]\( A \)[/tex] and [tex]\( B \)[/tex] with [tex]\( A = 0 \)[/tex] and [tex]\( B = 1 \)[/tex].
2. We print the initial values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. Thus, the first line of output is:
```
0 1
```
3. We then enter a `FOR` loop that iterates from [tex]\( I=1 \)[/tex] to [tex]\( I=5 \)[/tex]. During each iteration, the following steps take place:
- Calculate [tex]\( C = A + B \)[/tex]
- Print the new value of [tex]\( C \)[/tex]
- Update [tex]\( A \)[/tex] with the current value of [tex]\( B \)[/tex]
- Update [tex]\( B \)[/tex] with the current value of [tex]\( C \)[/tex]
Now, let's go through each iteration in detail.
Iteration 1 ([tex]\( I = 1 \)[/tex]):
- Initial values: [tex]\( A = 0 \)[/tex], [tex]\( B = 1 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 0 + 1 = 1 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
1
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 1 \)[/tex]
- [tex]\( B \leftarrow 1 \)[/tex]
Iteration 2 ([tex]\( I = 2 \)[/tex]):
- Initial values: [tex]\( A = 1 \)[/tex], [tex]\( B = 1 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 1 + 1 = 2 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
2
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 1 \)[/tex]
- [tex]\( B \leftarrow 2 \)[/tex]
Iteration 3 ([tex]\( I = 3 \)[/tex]):
- Initial values: [tex]\( A = 1 \)[/tex], [tex]\( B = 2 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 1 + 2 = 3 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
3
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 2 \)[/tex]
- [tex]\( B \leftarrow 3 \)[/tex]
Iteration 4 ([tex]\( I = 4 \)[/tex]):
- Initial values: [tex]\( A = 2 \)[/tex], [tex]\( B = 3 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 2 + 3 = 5 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
5
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 3 \)[/tex]
- [tex]\( B \leftarrow 5 \)[/tex]
Iteration 5 ([tex]\( I = 5 \)[/tex]):
- Initial values: [tex]\( A = 3 \)[/tex], [tex]\( B = 5 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 3 + 5 = 8 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
8
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 5 \)[/tex]
- [tex]\( B \leftarrow 8 \)[/tex]
Finally, combining all the printed outputs, we get:
```
0 1
1
2
3
5
8
```
This is the step-by-step output of the program.
1. We initialize two variables [tex]\( A \)[/tex] and [tex]\( B \)[/tex] with [tex]\( A = 0 \)[/tex] and [tex]\( B = 1 \)[/tex].
2. We print the initial values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. Thus, the first line of output is:
```
0 1
```
3. We then enter a `FOR` loop that iterates from [tex]\( I=1 \)[/tex] to [tex]\( I=5 \)[/tex]. During each iteration, the following steps take place:
- Calculate [tex]\( C = A + B \)[/tex]
- Print the new value of [tex]\( C \)[/tex]
- Update [tex]\( A \)[/tex] with the current value of [tex]\( B \)[/tex]
- Update [tex]\( B \)[/tex] with the current value of [tex]\( C \)[/tex]
Now, let's go through each iteration in detail.
Iteration 1 ([tex]\( I = 1 \)[/tex]):
- Initial values: [tex]\( A = 0 \)[/tex], [tex]\( B = 1 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 0 + 1 = 1 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
1
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 1 \)[/tex]
- [tex]\( B \leftarrow 1 \)[/tex]
Iteration 2 ([tex]\( I = 2 \)[/tex]):
- Initial values: [tex]\( A = 1 \)[/tex], [tex]\( B = 1 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 1 + 1 = 2 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
2
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 1 \)[/tex]
- [tex]\( B \leftarrow 2 \)[/tex]
Iteration 3 ([tex]\( I = 3 \)[/tex]):
- Initial values: [tex]\( A = 1 \)[/tex], [tex]\( B = 2 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 1 + 2 = 3 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
3
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 2 \)[/tex]
- [tex]\( B \leftarrow 3 \)[/tex]
Iteration 4 ([tex]\( I = 4 \)[/tex]):
- Initial values: [tex]\( A = 2 \)[/tex], [tex]\( B = 3 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 2 + 3 = 5 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
5
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 3 \)[/tex]
- [tex]\( B \leftarrow 5 \)[/tex]
Iteration 5 ([tex]\( I = 5 \)[/tex]):
- Initial values: [tex]\( A = 3 \)[/tex], [tex]\( B = 5 \)[/tex]
- Calculate [tex]\( C = A + B \Rightarrow C = 3 + 5 = 8 \)[/tex]
- Print [tex]\( C \)[/tex]:
```
8
```
- Update [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- [tex]\( A \leftarrow 5 \)[/tex]
- [tex]\( B \leftarrow 8 \)[/tex]
Finally, combining all the printed outputs, we get:
```
0 1
1
2
3
5
8
```
This is the step-by-step output of the program.