If p times the `p^(th)` term of an A.P is equal to q times its `q^(th)` term, then find `(p+q)^(th)` term of the A.P.

Options

A. `p+q^(2)`
B. `p-q^(2)`
C. `0`
D. `1`

Answer

To solve the problem step by step, we start with the given condition and then derive the required term of the arithmetic progression (A.P.).

### Step 1: Understand the terms of the A.P.
The general term of an A.P. can be expressed as:
\[ T_n = a + (n - 1)d \]
where \(T_n\) is the \(n^{th}\) term, \(a\) is the first term, and \(d\) is the common difference.

### Step 2: Write the expressions for the \(p^{th}\) and \(q^{th}\) terms.
Using the formula for the \(n^{th}\) term:
- The \(p^{th}\) term is:
\[ T_p = a + (p - 1)d \]
- The \(q^{th}\) term is:
\[ T_q = a + (q - 1)d \]

### Step 3: Set up the equation based on the problem statement.
According to the problem, we have:
\[ p \cdot T_p = q \cdot T_q \]
Substituting the expressions for \(T_p\) and \(T_q\):
\[ p \cdot (a + (p - 1)d) = q \cdot (a + (q - 1)d) \]

### Step 4: Expand both sides of the equation.
Expanding both sides gives:
\[ pa + p(p - 1)d = qa + q(q - 1)d \]

### Step 5: Rearrange the equation.
Rearranging the equation, we can group the terms involving \(a\) and \(d\):
\[ pa - qa = q(q - 1)d - p(p - 1)d \]
This simplifies to:
\[ (p - q)a = (q^2 - q - p^2 + p)d \]

### Step 6: Factor the right-hand side.
The right-hand side can be factored as:
\[ (p - q)a = (q^2 - p^2 + p - q)d \]
This can be rewritten using the difference of squares:
\[ (p - q)a = (q - p)(q + p + 1)d \]

### Step 7: Simplify the equation.
Since \(p - q\) is common on both sides, we can divide by \(p - q\) (assuming \(p \neq q\)):
\[ a = -(q + p + 1)d \]

### Step 8: Find the \((p + q)^{th}\) term.
Now, we need to find the \((p + q)^{th}\) term:
\[ T_{p + q} = a + (p + q - 1)d \]
Substituting \(a\) from the previous step:
\[ T_{p + q} = -(q + p + 1)d + (p + q - 1)d \]
This simplifies to:
\[ T_{p + q} = -d - d = -2d \]

### Final Answer:
Thus, the \((p + q)^{th}\) term of the A.P. is:
\[ T_{p + q} = -2d \]