(i) The sum of the first n terms of an AP is `((5n^(2))/(2) + (3n)/(2))`. Find the nth term and the 20th term of this AP.
(ii) The sum of the first n terms of an AP is `((3n^(2))/(2) + (5n)/(2)).` Find its nth term and the 25th term.

Answer

### Solution

#### Part (i)

1. **Given the sum of the first n terms of the AP:**
\[ S_n = \frac{5n^2}{2} + \frac{3n}{2} \]

2. **To find the first term \( a \):**
- Calculate \( S_1 \):
\[ S_1 = \frac{5(1)^2}{2} + \frac{3(1)}{2} = \frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4 \]
- Thus, the first term \( a = 4 \).

3. **To find the second term \( a_2 \):**
- Calculate \( S_2 \):
\[ S_2 = \frac{5(2)^2}{2} + \frac{3(2)}{2} = \frac{20}{2} + \frac{6}{2} = 10 + 3 = 13 \]
- The second term \( a_2 = S_2 - S_1 = 13 - 4 = 9 \).

4. **To find the third term \( a_3 \):**
- Calculate \( S_3 \):
\[ S_3 = \frac{5(3)^2}{2} + \frac{3(3)}{2} = \frac{45}{2} + \frac{9}{2} = \frac{54}{2} = 27 \]
- The third term \( a_3 = S_3 - S_2 = 27 - 13 = 14 \).

5. **Now we can find the common difference \( d \):**
\[ d = a_2 - a = 9 - 4 = 5 \]

6. **The nth term \( T_n \) of the AP is given by:**
\[ T_n = a + (n-1)d = 4 + (n-1) \cdot 5 = 4 + 5n - 5 = 5n - 1 \]

7. **To find the 20th term \( T_{20} \):**
\[ T_{20} = 5(20) - 1 = 100 - 1 = 99 \]

#### Summary for Part (i):
- The nth term is \( T_n = 5n - 1 \).
- The 20th term is \( T_{20} = 99 \).

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#### Part (ii)

1. **Given the sum of the first n terms of the AP:**
\[ S_n = \frac{3n^2}{2} + \frac{5n}{2} \]

2. **To find the first term \( a \):**
- Calculate \( S_1 \):
\[ S_1 = \frac{3(1)^2}{2} + \frac{5(1)}{2} = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 \]
- Thus, the first term \( a = 4 \).

3. **To find the second term \( a_2 \):**
- Calculate \( S_2 \):
\[ S_2 = \frac{3(2)^2}{2} + \frac{5(2)}{2} = \frac{12}{2} + \frac{10}{2} = 6 + 5 = 11 \]
- The second term \( a_2 = S_2 - S_1 = 11 - 4 = 7 \).

4. **To find the third term \( a_3 \):**
- Calculate \( S_3 \):
\[ S_3 = \frac{3(3)^2}{2} + \frac{5(3)}{2} = \frac{27}{2} + \frac{15}{2} = \frac{42}{2} = 21 \]
- The third term \( a_3 = S_3 - S_2 = 21 - 11 = 10 \).

5. **Now we can find the common difference \( d \):**
\[ d = a_2 - a = 7 - 4 = 3 \]

6. **The nth term \( T_n \) of the AP is given by:**
\[ T_n = a + (n-1)d = 4 + (n-1) \cdot 3 = 4 + 3n - 3 = 3n + 1 \]

7. **To find the 25th term \( T_{25} \):**
\[ T_{25} = 3(25) + 1 = 75 + 1 = 76 \]

#### Summary for Part (ii):
- The nth term is \( T_n = 3n + 1 \).
- The 25th term is \( T_{25} = 76 \).

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