If `p^("th"), 2p^("th") and 4p^("th")` terms of an arithmetic progression are in geometric progression, then the common ratio of the geometric progression is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, let's denote the terms of the arithmetic progression (AP) and analyze the conditions given in the problem.

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

For the terms given in the question:
- The \(p^{th}\) term is:
\[ T_p = a + (p-1)d \]
- The \(2p^{th}\) term is:
\[ T_{2p} = a + (2p-1)d \]
- The \(4p^{th}\) term is:
\[ T_{4p} = a + (4p-1)d \]

### Step 2: Set up the condition for GP
According to the problem, these three terms are in geometric progression (GP). For three terms \(x\), \(y\), and \(z\) to be in GP, the following condition must hold:
\[ y^2 = x \cdot z \]
In our case, we have:
\[ T_{2p}^2 = T_p \cdot T_{4p} \]

### Step 3: Substitute the terms
Substituting the expressions for \(T_p\), \(T_{2p}\), and \(T_{4p}\):
\[ (a + (2p-1)d)^2 = (a + (p-1)d)(a + (4p-1)d) \]

### Step 4: Expand both sides
Expanding the left side:
\[ (a + (2p-1)d)^2 = a^2 + 2a(2p-1)d + (2p-1)^2d^2 \]
Expanding the right side:
\[ (a + (p-1)d)(a + (4p-1)d) = a^2 + a(4p-1)d + a(p-1)d + (p-1)(4p-1)d^2 \]
This simplifies to:
\[ a^2 + (4p-1 + p-1)ad + (p-1)(4p-1)d^2 = a^2 + 3pad + (4p^2 - 5p + 1)d^2 \]

### Step 5: Set the expanded forms equal
Now, we set the two expansions equal to each other:
\[ a^2 + 2a(2p-1)d + (2p-1)^2d^2 = a^2 + 3pad + (4p^2 - 5p + 1)d^2 \]

### Step 6: Cancel \(a^2\) and rearrange
Cancelling \(a^2\) from both sides gives:
\[ 2a(2p-1)d + (2p-1)^2d^2 = 3pad + (4p^2 - 5p + 1)d^2 \]

### Step 7: Combine like terms
Rearranging gives:
\[ (2a(2p-1) - 3ap)d + ((2p-1)^2 - (4p^2 - 5p + 1))d^2 = 0 \]

### Step 8: Factor out \(d\)
Factoring out \(d\) (assuming \(d \neq 0\)):
\[ (2a(2p-1) - 3ap) + ((2p-1)^2 - (4p^2 - 5p + 1))d = 0 \]

### Step 9: Solve for the common ratio
From the GP condition:
\[ r = \frac{T_{2p}}{T_p} = \frac{a + (2p-1)d}{a + (p-1)d} \]
and
\[ r = \frac{T_{4p}}{T_{2p}} = \frac{a + (4p-1)d}{a + (2p-1)d} \]
Setting these equal gives us the common ratio.

### Final Answer
After simplifying, we find that the common ratio \(r = 2\).