Let `lambda` be the greatest integer for which `5p^(2)-16,2plambda,lambda^(2)` are jdistinct consecutive terms of an AP, where `p in R`. If the common difference of the Ap is `((m)/(n)),n in N` and m ,n are relative prime, the value of `m+n` is

Options

A. 133
B. 138
C. 143
D. 148

Explanation

To solve the problem step by step, we need to find the greatest integer `lambda` such that the terms `5p^2 - 16`, `2p*lambda`, and `lambda^2` are distinct consecutive terms of an arithmetic progression (AP). We will also find the common difference of the AP and express it in the form \( \frac{m}{n} \), where \( m \) and \( n \) are coprime integers, and finally compute \( m + n \).

### Step 1: Set up the equation for the AP
The terms are in AP if:
\[ 2(2p\lambda) = (5p^2 - 16) + (\lambda^2) \]
This simplifies to:
\[ 4p\lambda = 5p^2 - 16 + \lambda^2 \]
Rearranging gives us:
\[ 5p^2 - 4p\lambda + \lambda^2 - 16 = 0 \]

### Step 2: Determine the discriminant
For the quadratic equation in \( p \) to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = (-4\lambda)^2 - 4 \cdot 5 \cdot (\lambda^2 - 16) \geq 0 \]
Calculating the discriminant:
\[ D = 16\lambda^2 - 20(\lambda^2 - 16) = 16\lambda^2 - 20\lambda^2 + 320 = -4\lambda^2 + 320 \]
Setting the discriminant greater than or equal to zero:
\[ -4\lambda^2 + 320 \geq 0 \implies 4\lambda^2 \leq 320 \implies \lambda^2 \leq 80 \]
Thus:
\[ \lambda \leq \sqrt{80} = 4\sqrt{5} \approx 8.944 \]

### Step 3: Find the greatest integer value of `lambda`
The greatest integer less than or equal to \( 4\sqrt{5} \) is \( 8 \). Therefore:
\[ \lambda = 8 \]

### Step 4: Substitute `lambda` back into the equation
Substituting \( \lambda = 8 \) into the quadratic equation:
\[ 5p^2 - 4(8)p + (8^2 - 16) = 0 \implies 5p^2 - 32p + 48 = 0 \]

### Step 5: Factor the quadratic equation
We can factor this equation:
\[ D = (-32)^2 - 4 \cdot 5 \cdot 48 = 1024 - 960 = 64 \]
The roots are:
\[ p = \frac{32 \pm \sqrt{64}}{2 \cdot 5} = \frac{32 \pm 8}{10} = \frac{40}{10} \text{ or } \frac{24}{10} \implies p = 4 \text{ or } p = \frac{12}{5} \]

### Step 6: Determine valid `p`
Since \( p = 4 \) would make the terms not distinct (as \( 2p\lambda = \lambda^2 \)), we take:
\[ p = \frac{12}{5} \]

### Step 7: Calculate the common difference
Now, we find the common difference \( d \):
\[ d = 2p\lambda - (5p^2 - 16) = \lambda^2 - 2p\lambda \]
Calculating \( d \):
\[ d = 8^2 - 2 \cdot \frac{12}{5} \cdot 8 = 64 - \frac{192}{5} = \frac{320 - 192}{5} = \frac{128}{5} \]

### Step 8: Express \( d \) in the form \( \frac{m}{n} \)
Here, \( m = 128 \) and \( n = 5 \). Since \( 128 \) and \( 5 \) are coprime, we can find:
\[ m + n = 128 + 5 = 133 \]

### Final Answer
Thus, the value of \( m + n \) is:
\[ \boxed{133} \]