(i) The 3rd term of an A.P. is 1 and 6 th term is -11. Determine its 15th term and rth term.
(ii) In an A.P. , the third term is p and the fourth term is q, find the 10 th term and the general term.

Answer

Let's solve the problem step by step.

### Part (i)

**Given:**
- The 3rd term of an A.P. (Arithmetic Progression) is 1.
- The 6th term of the A.P. is -11.

**Step 1: Write the formulas for the terms.**
- The nth term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

**Step 2: Set up equations for the given terms.**
- For the 3rd term:
\[ T_3 = a + 2d = 1 \quad \text{(Equation 1)} \]
- For the 6th term:
\[ T_6 = a + 5d = -11 \quad \text{(Equation 2)} \]

**Step 3: Subtract Equation 1 from Equation 2.**
- This will eliminate \( a \):
\[ (a + 5d) - (a + 2d) = -11 - 1 \]
Simplifying gives:
\[ 3d = -12 \]
Thus,
\[ d = -4 \]

**Step 4: Substitute \( d \) back into Equation 1 to find \( a \).**
- Substitute \( d = -4 \) into Equation 1:
\[ a + 2(-4) = 1 \]
Simplifying gives:
\[ a - 8 = 1 \implies a = 9 \]

**Step 5: Find the 15th term.**
- Using the formula for the nth term:
\[ T_{15} = a + 14d \]
Substituting \( a = 9 \) and \( d = -4 \):
\[ T_{15} = 9 + 14(-4) = 9 - 56 = -47 \]

**Step 6: Find the r-th term.**
- The r-th term can be expressed as:
\[ T_r = a + (r-1)d \]
Substituting \( a = 9 \) and \( d = -4 \):
\[ T_r = 9 + (r-1)(-4) = 9 - 4(r-1) = 9 - 4r + 4 = 13 - 4r \]

### Final Answers for Part (i):
- 15th term: \( T_{15} = -47 \)
- r-th term: \( T_r = 13 - 4r \)

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### Part (ii)

**Given:**
- The 3rd term of an A.P. is \( p \).
- The 4th term of the A.P. is \( q \).

**Step 1: Write the equations for the given terms.**
- For the 3rd term:
\[ T_3 = a + 2d = p \quad \text{(Equation 1)} \]
- For the 4th term:
\[ T_4 = a + 3d = q \quad \text{(Equation 2)} \]

**Step 2: Subtract Equation 1 from Equation 2.**
- This will eliminate \( a \):
\[ (a + 3d) - (a + 2d) = q - p \]
Simplifying gives:
\[ d = q - p \]

**Step 3: Substitute \( d \) back into Equation 1 to find \( a \).**
- Substitute \( d = q - p \) into Equation 1:
\[ a + 2(q - p) = p \]
Simplifying gives:
\[ a + 2q - 2p = p \implies a = p - 2q + 2p = 3p - 2q \]

**Step 4: Find the 10th term.**
- Using the formula for the nth term:
\[ T_{10} = a + 9d \]
Substituting \( a = 3p - 2q \) and \( d = q - p \):
\[ T_{10} = (3p - 2q) + 9(q - p) = 3p - 2q + 9q - 9p = -6p + 7q \]

**Step 5: Write the general term.**
- The general term \( T_n \) can be expressed as:
\[ T_n = a + (n-1)d \]
Substituting \( a = 3p - 2q \) and \( d = q - p \):
\[ T_n = (3p - 2q) + (n-1)(q - p) \]
Expanding gives:
\[ T_n = 3p - 2q + (n-1)q - (n-1)p = (3 - n + 1)p + (-2 + n - 1)q \]
Simplifying gives:
\[ T_n = (4 - n)p + (n - 3)q \]

### Final Answers for Part (ii):
- 10th term: \( T_{10} = -6p + 7q \)
- General term: \( T_n = (4 - n)p + (n - 3)q \)

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