If p times the pth term of an AP is q times the qth term, then what is the `(p+q)th` term equal to ?

Options

A. `p+q`
B. pq
C. 1
D. 0

Explanation

To solve the problem, we need to find the \((p+q)\)th term of an arithmetic progression (AP) given that \(p\) times the \(p\)th term is equal to \(q\) times the \(q\)th term. Let's go through the solution step-by-step.

### Step 1: Define the \(n\)th term of an AP
The \(n\)th term of an arithmetic progression can be expressed as:
\[ T_n = a + (n-1)d \]
where \(a\) is the first term and \(d\) is the common difference.

### Step 2: Write expressions for the \(p\)th and \(q\)th terms
Using the formula for the \(n\)th term, we can write:
\[ T_p = a + (p-1)d \]
\[ T_q = a + (q-1)d \]

### Step 3: Set up the given condition
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
Expanding both sides gives:
\[ p \cdot a + p(p-1)d = q \cdot a + q(q-1)d \]

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

### Step 6: Factor the right side
We can factor the right side:
\[ (p - q)a = (q^2 - p^2 + p - q)d \]
Using the difference of squares:
\[ (p - q)a = (q - p)(q + p + 1)d \]

### Step 7: Divide both sides by \(p - q\) (assuming \(p \neq q\))
Dividing both sides by \(p - q\) gives:
\[ a = -(q + p + 1)d \]

### 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 the value of \(a\):
\[ T_{p+q} = -(q + p + 1)d + (p + q - 1)d \]
Simplifying this:
\[ T_{p+q} = -(q + p + 1)d + (p + q - 1)d \]
\[ = -qd - pd - d + pd + qd - d \]
\[ = -d \]

### Final Answer
Thus, the \((p + q)\)th term of the AP is:
\[ T_{p+q} = 0 \]