In an arithmetic progression the `(p+1)^("th")` term is twice the `(q+1)^("th")` term. If its `(3p+1)^("th")` term is `lambda` times the`(p+q+1)^("th")` term, then `lambda` is equal to

Options

A. 2
B. `(1)/(2)`
C. 3
D. `(1)/(3)`

Answer

To solve the problem, we need to analyze the conditions given in the arithmetic progression (AP) and derive the value of \( \lambda \).

### Step 1: Understanding the nth term of an AP
The nth 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: Setting up the equations based on the problem statement
According to the problem:
1. The \( (p + 1)^{th} \) term is twice the \( (q + 1)^{th} \) term:
\[ T_{p+1} = 2T_{q+1} \]
Substituting the formula for the nth term:
\[ a + pd = 2(a + qd) \]

### Step 3: Simplifying the first equation
Expanding the equation:
\[ a + pd = 2a + 2qd \]
Rearranging gives:
\[ pd - 2qd = 2a - a \]
\[ pd - 2qd = a \]
Thus, we can express \( a \) in terms of \( p \), \( q \), and \( d \):
\[ a = pd - 2qd \]

### Step 4: Setting up the second equation
2. The \( (3p + 1)^{th} \) term is \( \lambda \) times the \( (p + q + 1)^{th} \) term:
\[ T_{3p+1} = \lambda T_{p+q+1} \]
Substituting the nth term formula:
\[ a + 3pd = \lambda (a + (p + q)d) \]

### Step 5: Simplifying the second equation
Expanding the equation:
\[ a + 3pd = \lambda a + \lambda (p + q)d \]
Rearranging gives:
\[ a + 3pd - \lambda a - \lambda (p + q)d = 0 \]
\[ (1 - \lambda)a + (3p - \lambda(p + q))d = 0 \]

### Step 6: Substituting the value of \( a \)
Now, substitute \( a = pd - 2qd \) into the equation:
\[ (1 - \lambda)(pd - 2qd) + (3p - \lambda(p + q))d = 0 \]
Distributing gives:
\[ (1 - \lambda)pd - 2(1 - \lambda)qd + (3p - \lambda(p + q))d = 0 \]

### Step 7: Collecting terms
Combining the terms:
\[ [(1 - \lambda)p + (3p - \lambda(p + q))]d - 2(1 - \lambda)qd = 0 \]
This must hold for \( d \neq 0 \), leading to:
\[ (1 - \lambda)p + 3p - \lambda(p + q) = 2(1 - \lambda)q \]

### Step 8: Solving for \( \lambda \)
Rearranging gives:
\[ (1 - \lambda)p + 3p - \lambda p - \lambda q = 2(1 - \lambda)q \]
Combining like terms:
\[ (1 - \lambda + 3 - \lambda)p = (2(1 - \lambda) + \lambda)q \]
This simplifies to:
\[ (4 - 2\lambda)p = (2 - \lambda)q \]

### Step 9: Finding \( \lambda \)
To find \( \lambda \), we can equate the coefficients:
\[ \lambda = 2 \]

Thus, the value of \( \lambda \) is:
\[ \lambda = 2 \]