Find the terms indicated in each case:
(i) `a_(n)=4n-3,a_(17),a_(24)`
(ii) `a_(n)=(n-1)(2-n)(3+n),a_(1),a_(2),a_(3)`

Answer

Let's solve the given problem step by step.

### Part (i): Find \( a_{17} \) and \( a_{24} \) for \( a_n = 4n - 3 \)

1. **Finding \( a_{17} \)**:
- Substitute \( n = 17 \) into the formula:
\[ a_{17} = 4(17) - 3 \]
- Calculate:
\[ a_{17} = 68 - 3 = 65 \]

2. **Finding \( a_{24} \)**:
- Substitute \( n = 24 \) into the formula:
\[ a_{24} = 4(24) - 3 \]
- Calculate:
\[ a_{24} = 96 - 3 = 93 \]

### Results for Part (i):
- \( a_{17} = 65 \)
- \( a_{24} = 93 \)

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### Part (ii): Find \( a_1 \), \( a_2 \), and \( a_3 \) for \( a_n = (n-1)(2-n)(3+n) \)

1. **Finding \( a_1 \)**:
- Substitute \( n = 1 \) into the formula:
\[ a_1 = (1-1)(2-1)(3+1) \]
- Calculate:
\[ a_1 = 0 \cdot 1 \cdot 4 = 0 \]

2. **Finding \( a_2 \)**:
- Substitute \( n = 2 \) into the formula:
\[ a_2 = (2-1)(2-2)(3+2) \]
- Calculate:
\[ a_2 = 1 \cdot 0 \cdot 5 = 0 \]

3. **Finding \( a_3 \)**:
- Substitute \( n = 3 \) into the formula:
\[ a_3 = (3-1)(2-3)(3+3) \]
- Calculate:
\[ a_3 = 2 \cdot (-1) \cdot 6 = -12 \]

### Results for Part (ii):
- \( a_1 = 0 \)
- \( a_2 = 0 \)
- \( a_3 = -12 \)

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### Summary of Results:
- For part (i): \( a_{17} = 65 \), \( a_{24} = 93 \)
- For part (ii): \( a_1 = 0 \), \( a_2 = 0 \), \( a_3 = -12 \)

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