There are two sets A and B each of which consists of three numbers in AP whose sum is 15 and where D and d are the common differences such that `D=1+d,dgt0.` If p=7(q-p)`, where p and q are the product of the nymbers respectively in the two series.
The value of p is

Options

A. 105
B. 140
C. 175
D. 210

Explanation

To solve the problem step by step, we will break down the information given and derive the required value of \( p \).

### Step 1: Understanding the Sets
Let the three numbers in set \( A \) be \( a - D, a, a + D \) and the three numbers in set \( B \) be \( b - d, b, b + d \). Since both sets are in arithmetic progression (AP) and their sums equal 15, we can write:

\[ (a - D) + a + (a + D) = 15 \implies 3a = 15 \implies a = 5 \]

Similarly for set \( B \):

\[ (b - d) + b + (b + d) = 15 \implies 3b = 15 \implies b = 5 \]

### Step 2: Finding the Common Differences
We know from the problem that \( D = 1 + d \) and \( d > 0 \).

### Step 3: Finding the Products
The product \( p \) for set \( A \) can be calculated as:

\[ p = (a - D) \cdot a \cdot (a + D) = (5 - D) \cdot 5 \cdot (5 + D) \]

This can be simplified using the identity \( (x - y)(x)(x + y) = x^3 - y^2 \):

\[ p = 5 \cdot (5^2 - D^2) = 5 \cdot (25 - D^2) \]

For set \( B \):

\[ q = (b - d) \cdot b \cdot (b + d) = (5 - d) \cdot 5 \cdot (5 + d) \]

Similarly, we can simplify this:

\[ q = 5 \cdot (5^2 - d^2) = 5 \cdot (25 - d^2) \]

### Step 4: Using the Given Relationship
We are given that:

\[ p = 7(q - p) \]

Rearranging gives:

\[ p + 7p = 7q \implies 8p = 7q \implies \frac{p}{q} = \frac{7}{8} \]

### Step 5: Setting Up the Ratio
Now substituting the expressions for \( p \) and \( q \):

\[ \frac{5(25 - D^2)}{5(25 - d^2)} = \frac{7}{8} \]

This simplifies to:

\[ \frac{25 - D^2}{25 - d^2} = \frac{7}{8} \]

Cross-multiplying gives:

\[ 8(25 - D^2) = 7(25 - d^2) \]

Expanding both sides:

\[ 200 - 8D^2 = 175 - 7d^2 \]

Rearranging leads to:

\[ 25 = 8D^2 - 7d^2 \]

### Step 6: Substituting \( D \)
Since \( D = 1 + d \), we can substitute \( D \) in terms of \( d \):

\[ 25 = 8(1 + d)^2 - 7d^2 \]

Expanding \( (1 + d)^2 \):

\[ = 8(1 + 2d + d^2) - 7d^2 = 8 + 16d + 8d^2 - 7d^2 \]

This simplifies to:

\[ 25 = 8 + 16d + d^2 \]

Rearranging gives:

\[ d^2 + 16d - 17 = 0 \]

### Step 7: Solving the Quadratic
Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ d = \frac{-16 \pm \sqrt{16^2 + 4 \cdot 17}}{2} = \frac{-16 \pm \sqrt{256 + 68}}{2} = \frac{-16 \pm \sqrt{324}}{2} = \frac{-16 \pm 18}{2} \]

Calculating the two possible values:

1. \( d = \frac{2}{2} = 1 \) (valid since \( d > 0 \))
2. \( d = \frac{-34}{2} = -17 \) (not valid)

### Step 8: Finding \( D \) and \( p \)
Now substituting \( d = 1 \):

\[ D = 1 + d = 2 \]

Finally, substituting \( D \) back into the expression for \( p \):

\[ p = 5(25 - D^2) = 5(25 - 4) = 5 \cdot 21 = 105 \]

Thus, the value of \( p \) is:

\[ \boxed{105} \]