In an A.P, the first term is 1 and sum of the first p terms is 0, then sum of the first (p + q) terms is

Options

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

Answer

To solve the problem step by step, we will use the properties of arithmetic progressions (A.P.) and the formula for the sum of the first n terms of an A.P.

### Step 1: Understand the given information
We know that:
- The first term \( a = 1 \)
- The sum of the first \( p \) terms \( S_p = 0 \)

### Step 2: Use the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 3: Set up the equation for \( S_p \)
Substituting \( n = p \) and \( a = 1 \) into the formula, we have:
\[ S_p = \frac{p}{2} \left(2 \cdot 1 + (p - 1)d\right) = 0 \]
This simplifies to:
\[ S_p = \frac{p}{2} \left(2 + (p - 1)d\right) = 0 \]

### Step 4: Analyze the equation
Since \( p \neq 0 \) (as it represents the number of terms), we can conclude that:
\[ 2 + (p - 1)d = 0 \]
From this, we can isolate \( d \):
\[ (p - 1)d = -2 \quad \Rightarrow \quad d = \frac{-2}{p - 1} \]

### Step 5: Find the sum of the first \( p + q \) terms
Now, we need to find \( S_{p+q} \):
\[ S_{p+q} = \frac{p + q}{2} \left(2a + (p + q - 1)d\right) \]
Substituting \( a = 1 \):
\[ S_{p+q} = \frac{p + q}{2} \left(2 + (p + q - 1)d\right) \]

### Step 6: Substitute \( d \) into the equation
Now substitute \( d = \frac{-2}{p - 1} \):
\[ S_{p+q} = \frac{p + q}{2} \left(2 + (p + q - 1) \left(\frac{-2}{p - 1}\right)\right) \]
This simplifies to:
\[ S_{p+q} = \frac{p + q}{2} \left(2 - \frac{2(p + q - 1)}{p - 1}\right) \]

### Step 7: Simplify the expression
Combining the terms inside the parentheses:
\[ S_{p+q} = \frac{p + q}{2} \left(\frac{2(p - 1) - 2(p + q - 1)}{p - 1}\right) \]
This simplifies to:
\[ S_{p+q} = \frac{p + q}{2} \left(\frac{2p - 2 - 2p - 2q + 2}{p - 1}\right) \]
\[ = \frac{p + q}{2} \left(\frac{-2q}{p - 1}\right) \]
\[ = \frac{-(p + q)q}{p - 1} \]

### Final Result
Thus, the sum of the first \( p + q \) terms is:
\[ S_{p+q} = \frac{-(p + q)q}{p - 1} \]