Answer :
Let's solve the given questions one by one with detailed steps.
### Question 9: Sequence 3, 3, 9, 6, 15, 12, ...
#### (a) Calculate [tex]\( I_{52} - I_{51} \)[/tex]
1. The sequence alternates between two sub-sequences:
- Odd-indexed terms: [tex]\(3, 9, 15, \dots\)[/tex]
- Even-indexed terms: [tex]\(3, 6, 12, \dots\)[/tex]
2. The odd-indexed terms form an arithmetic sequence with the first term 3 and common difference of 6:
- General formula for odd-indexed terms: [tex]\( I_{2m-1} = 3 + 6 \times (m-1) \)[/tex]
3. The even-indexed terms also form an arithmetic sequence with the first term 3 and common difference of 6:
- General formula for even-indexed terms: [tex]\( I_{2m} = 3 + 6 \times (m-1) \)[/tex]
4. Determine the indices:
- [tex]\( I_{51} \)[/tex] is odd-indexed, so use the formula for odd-indexed terms with [tex]\( m = 26 \)[/tex].
- [tex]\( I_{52} \)[/tex] is even-indexed, so use the formula for even-indexed terms with [tex]\( m = 26 \)[/tex].
5. Calculate:
- [tex]\( I_{51} = 3 + 6 \times (26-1) = 3 + 6 \times 25 = 3 + 150 = 153 \)[/tex]
- [tex]\( I_{52} = 3 + 6 \times (26-1) = 3 + 6 \times 25 = 3 + 150 = 153 \)[/tex]
6. Therefore, [tex]\( I_{52} - I_{51} = 153 - 153 = 0 \)[/tex].
#### (b) Will this sequence always be divisible by 3? Motivate.
1. Both subsequences are arithmetic sequences with a common difference of 6.
2. Since the common difference, 6, is a multiple of 3, and both sequences start with 3 (which is divisible by 3), every term in both sequences will also be divisible by 3.
3. Therefore, the entire sequence will always be divisible by 3.
#### (c) Calculate the sum of the first 50 terms of this sequence.
1. The first 50 terms are comprised of 25 odd-indexed terms and 25 even-indexed terms.
2. For the odd-indexed subsequence ([tex]\(a = 3\)[/tex], [tex]\(d = 6\)[/tex], [tex]\(n = 25\)[/tex]):
- The sum of the first 25 odd-indexed terms, [tex]\( S_{odd} \)[/tex], can be calculated using the sum formula for an arithmetic sequence:
[tex]\[ S_{odd} = \frac{n}{2} (2a + (n-1)d) = \frac{25}{2} (2 \times 3 + (25-1) \times 6) = \frac{25}{2} (6 + 144) = \frac{25}{2} \times 150 = 1875 \][/tex]
3. For the even-indexed subsequence ([tex]\(a = 3\)[/tex], [tex]\(d = 6\)[/tex], [tex]\(n = 25\)[/tex]):
- The sum of the first 25 even-indexed terms, [tex]\( S_{even} \)[/tex], can be calculated similarly:
[tex]\[ S_{even} = \frac{n}{2} (2a + (n-1)d) = \frac{25}{2} (2 \times 3 + (25-1) \times 6) = \frac{25}{2} (6 + 144) = \frac{25}{2} \times 150 = 1875 \][/tex]
4. Total sum of the first 50 terms, [tex]\( S_{total} \)[/tex], is the sum of [tex]\( S_{odd} \)[/tex] and [tex]\( S_{even} \)[/tex]:
[tex]\[ S_{total} = S_{odd} + S_{even} = 1875 + 1875 = 3750 \][/tex]
### Summary of results:
1. [tex]\( I_{52} - I_{51} = \boxed{0} \)[/tex]
2. The sequence will always be divisible by 3.
3. The sum of the first 50 terms of the sequence is [tex]\( \boxed{3750} \)[/tex].
### Question 9: Sequence 3, 3, 9, 6, 15, 12, ...
#### (a) Calculate [tex]\( I_{52} - I_{51} \)[/tex]
1. The sequence alternates between two sub-sequences:
- Odd-indexed terms: [tex]\(3, 9, 15, \dots\)[/tex]
- Even-indexed terms: [tex]\(3, 6, 12, \dots\)[/tex]
2. The odd-indexed terms form an arithmetic sequence with the first term 3 and common difference of 6:
- General formula for odd-indexed terms: [tex]\( I_{2m-1} = 3 + 6 \times (m-1) \)[/tex]
3. The even-indexed terms also form an arithmetic sequence with the first term 3 and common difference of 6:
- General formula for even-indexed terms: [tex]\( I_{2m} = 3 + 6 \times (m-1) \)[/tex]
4. Determine the indices:
- [tex]\( I_{51} \)[/tex] is odd-indexed, so use the formula for odd-indexed terms with [tex]\( m = 26 \)[/tex].
- [tex]\( I_{52} \)[/tex] is even-indexed, so use the formula for even-indexed terms with [tex]\( m = 26 \)[/tex].
5. Calculate:
- [tex]\( I_{51} = 3 + 6 \times (26-1) = 3 + 6 \times 25 = 3 + 150 = 153 \)[/tex]
- [tex]\( I_{52} = 3 + 6 \times (26-1) = 3 + 6 \times 25 = 3 + 150 = 153 \)[/tex]
6. Therefore, [tex]\( I_{52} - I_{51} = 153 - 153 = 0 \)[/tex].
#### (b) Will this sequence always be divisible by 3? Motivate.
1. Both subsequences are arithmetic sequences with a common difference of 6.
2. Since the common difference, 6, is a multiple of 3, and both sequences start with 3 (which is divisible by 3), every term in both sequences will also be divisible by 3.
3. Therefore, the entire sequence will always be divisible by 3.
#### (c) Calculate the sum of the first 50 terms of this sequence.
1. The first 50 terms are comprised of 25 odd-indexed terms and 25 even-indexed terms.
2. For the odd-indexed subsequence ([tex]\(a = 3\)[/tex], [tex]\(d = 6\)[/tex], [tex]\(n = 25\)[/tex]):
- The sum of the first 25 odd-indexed terms, [tex]\( S_{odd} \)[/tex], can be calculated using the sum formula for an arithmetic sequence:
[tex]\[ S_{odd} = \frac{n}{2} (2a + (n-1)d) = \frac{25}{2} (2 \times 3 + (25-1) \times 6) = \frac{25}{2} (6 + 144) = \frac{25}{2} \times 150 = 1875 \][/tex]
3. For the even-indexed subsequence ([tex]\(a = 3\)[/tex], [tex]\(d = 6\)[/tex], [tex]\(n = 25\)[/tex]):
- The sum of the first 25 even-indexed terms, [tex]\( S_{even} \)[/tex], can be calculated similarly:
[tex]\[ S_{even} = \frac{n}{2} (2a + (n-1)d) = \frac{25}{2} (2 \times 3 + (25-1) \times 6) = \frac{25}{2} (6 + 144) = \frac{25}{2} \times 150 = 1875 \][/tex]
4. Total sum of the first 50 terms, [tex]\( S_{total} \)[/tex], is the sum of [tex]\( S_{odd} \)[/tex] and [tex]\( S_{even} \)[/tex]:
[tex]\[ S_{total} = S_{odd} + S_{even} = 1875 + 1875 = 3750 \][/tex]
### Summary of results:
1. [tex]\( I_{52} - I_{51} = \boxed{0} \)[/tex]
2. The sequence will always be divisible by 3.
3. The sum of the first 50 terms of the sequence is [tex]\( \boxed{3750} \)[/tex].
