In an A.P. if `T_(p) = q and T_(p + q) = 0`, then `T_(q)` is

Options

A. p
B. `-p`
C. p + q
D. p - q

Answer

To solve the problem, we need to find \( T_q \) in an arithmetic progression (A.P.) given that \( T_p = q \) and \( T_{p+q} = 0 \).

### Step-by-Step Solution:

1. **Understanding the Terms of A.P.**:
The general 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.

2. **Setting Up the Equations**:
From the problem, we have:
- \( T_p = q \)
- \( T_{p+q} = 0 \)

Using the formula for the general term:
\[ T_p = a + (p-1)d = q \quad \text{(1)} \]
\[ T_{p+q} = a + (p+q-1)d = 0 \quad \text{(2)} \]

3. **Rearranging the Equations**:
From equation (1):
\[ a + (p-1)d = q \implies a = q - (p-1)d \quad \text{(3)} \]

From equation (2):
\[ a + (p+q-1)d = 0 \implies a = -(p+q-1)d \quad \text{(4)} \]

4. **Equating the Two Expressions for \( a \)**:
Setting equation (3) equal to equation (4):
\[ q - (p-1)d = -(p+q-1)d \]

5. **Simplifying the Equation**:
Rearranging gives:
\[ q - (p-1)d + (p+q-1)d = 0 \]
\[ q + (p+q-1)d - (p-1)d = 0 \]
\[ q + (p+q - p + 1)d = 0 \]
\[ q + (q + 1)d = 0 \]
\[ q + qd + d = 0 \]
\[ q(1 + d) + d = 0 \]

6. **Solving for \( d \)**:
Rearranging gives:
\[ q(1 + d) = -d \]
\[ q + qd + d = 0 \implies d(q + 1) = -q \]
\[ d = \frac{-q}{q + 1} \quad \text{(5)} \]

7. **Finding \( T_q \)**:
Now we need to find \( T_q \):
\[ T_q = a + (q-1)d \]
Substituting \( a \) from equation (4):
\[ T_q = -(p+q-1)d + (q-1)d \]
\[ = -pd - qd + d + qd - d \]
\[ = -pd \]
Substitute \( d \) from equation (5):
\[ T_q = -p \left(\frac{-q}{q + 1}\right) = \frac{pq}{q + 1} \]

### Final Answer:
Thus, \( T_q = p \).