If sum of an indinite `G.P p,1,1//p,1//p^2`…=9/2.. Is then value of p is

Options

A. 2
B. `3//2`
C. 3
D. `9//2`

Explanation

To find the value of \( p \) given that the sum of the infinite geometric progression (G.P) \( p, 1, \frac{1}{p}, \frac{1}{p^2}, \ldots \) equals \( \frac{9}{2} \), we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \( a \) of the G.P is \( p \). The second term is \( 1 \), and the third term is \( \frac{1}{p} \).

To find the common ratio \( r \):
\[ r = \frac{\text{second term}}{\text{first term}} = \frac{1}{p} \]

### Step 2: Write the formula for the sum of an infinite G.P
The sum \( S \) of an infinite G.P can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values of \( a \) and \( r \):
\[ S = \frac{p}{1 - \frac{1}{p}} \]

### Step 3: Simplify the expression for the sum
Simplifying the denominator:
\[ 1 - \frac{1}{p} = \frac{p - 1}{p} \]
Thus, the sum becomes:
\[ S = \frac{p}{\frac{p - 1}{p}} = \frac{p^2}{p - 1} \]

### Step 4: Set the sum equal to \( \frac{9}{2} \)
According to the problem, the sum equals \( \frac{9}{2} \):
\[ \frac{p^2}{p - 1} = \frac{9}{2} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 2p^2 = 9(p - 1) \]
Expanding the right side:
\[ 2p^2 = 9p - 9 \]

### Step 6: Rearrange the equation to form a quadratic equation
Rearranging gives:
\[ 2p^2 - 9p + 9 = 0 \]

### Step 7: Factor the quadratic equation
To factor \( 2p^2 - 9p + 9 \), we look for two numbers that multiply to \( 2 \times 9 = 18 \) and add to \( -9 \). The numbers are \( -6 \) and \( -3 \):
\[ 2p^2 - 6p - 3p + 9 = 0 \]
Grouping gives:
\[ 2p(p - 3) - 3(p - 3) = 0 \]
Factoring out \( (p - 3) \):
\[ (2p - 3)(p - 3) = 0 \]

### Step 8: Solve for \( p \)
Setting each factor to zero gives:
1. \( 2p - 3 = 0 \) → \( p = \frac{3}{2} \)
2. \( p - 3 = 0 \) → \( p = 3 \)

Thus, the values of \( p \) are \( \frac{3}{2} \) and \( 3 \).

### Final Answer
The values of \( p \) are \( \frac{3}{2} \) and \( 3 \).

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