There are two sets `M_1` and `M_2` each of which consists of three numbers in arithmetic sequence whose sum is 15. Let `d_1` and `d_2` be the common differences such that `d_1=1+d_2` and `8p_1=7p_2` where `p_1` and `p_2` are the product of the numbers respectively in `M_1` and `M_2`. If `d_2 gt 0` then find the value of `(p_2-p_1)/(d_1+d_2)`

Options

A. 108
B. 120
C. 144
D. 160

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the sets M1 and M2
Let the first set \( M_1 \) consist of three numbers in arithmetic progression:
\[ M_1 = \{ a_1 - d_1, a_1, a_1 + d_1 \} \]
The sum of the numbers in \( M_1 \) is:
\[ (a_1 - d_1) + a_1 + (a_1 + d_1) = 3a_1 = 15 \]
From this, we find:
\[ a_1 = 5 \]

### Step 2: Define the second set M2
Similarly, let the second set \( M_2 \) consist of three numbers in arithmetic progression:
\[ M_2 = \{ a_2 - d_2, a_2, a_2 + d_2 \} \]
The sum of the numbers in \( M_2 \) is:
\[ (a_2 - d_2) + a_2 + (a_2 + d_2) = 3a_2 = 15 \]
From this, we find:
\[ a_2 = 5 \]

### Step 3: Express the sets using common differences
Now we can express the sets using the common differences:
\[ M_1 = \{ 5 - d_1, 5, 5 + d_1 \} \]
\[ M_2 = \{ 5 - d_2, 5, 5 + d_2 \} \]

### Step 4: Relate the common differences
Given that \( d_1 = 1 + d_2 \), we can express \( d_2 \) in terms of \( d_1 \):
\[ d_2 = d_1 - 1 \]

### Step 5: Calculate the products \( p_1 \) and \( p_2 \)
The product \( p_1 \) for set \( M_1 \) is:
\[ p_1 = (5 - d_1) \cdot 5 \cdot (5 + d_1) = 5(25 - d_1^2) \]
The product \( p_2 \) for set \( M_2 \) is:
\[ p_2 = (5 - d_2) \cdot 5 \cdot (5 + d_2) = 5(25 - d_2^2) \]

### Step 6: Substitute \( d_2 \) into \( p_2 \)
Substituting \( d_2 = d_1 - 1 \) into \( p_2 \):
\[ p_2 = 5(25 - (d_1 - 1)^2) = 5(25 - (d_1^2 - 2d_1 + 1)) = 5(24 + 2d_1 - d_1^2) \]

### Step 7: Use the given relationship \( 8p_1 = 7p_2 \)
Substituting \( p_1 \) and \( p_2 \) into the equation:
\[ 8 \cdot 5(25 - d_1^2) = 7 \cdot 5(24 + 2d_1 - d_1^2) \]
Dividing both sides by 5:
\[ 8(25 - d_1^2) = 7(24 + 2d_1 - d_1^2) \]
Expanding both sides:
\[ 200 - 8d_1^2 = 168 + 14d_1 - 7d_1^2 \]
Rearranging gives:
\[ d_1^2 + 14d_1 - 32 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[ d_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-14 \pm \sqrt{196 + 128}}{2} = \frac{-14 \pm 18}{2} \]
Calculating the roots:
\[ d_1 = 2 \quad \text{(since \( d_1 > 0 \))} \]
Then, substituting back to find \( d_2 \):
\[ d_2 = d_1 - 1 = 1 \]

### Step 9: Calculate \( p_1 \) and \( p_2 \)
Now we can find \( p_1 \) and \( p_2 \):
\[ p_1 = 5(25 - 2^2) = 5(25 - 4) = 5 \cdot 21 = 105 \]
\[ p_2 = 5(24 + 2 \cdot 2 - 2^2) = 5(24 + 4 - 4) = 5 \cdot 24 = 120 \]

### Step 10: Calculate \( \frac{p_2 - p_1}{d_1 + d_2} \)
Finally, we calculate:
\[ \frac{p_2 - p_1}{d_1 + d_2} = \frac{120 - 105}{2 + 1} = \frac{15}{3} = 5 \]

Thus, the final answer is:
\[ \boxed{5} \]