one `AM` ,`a` and two `GM's` ,` p `and `q` be inserted between any two given numbers then show that `p^3+q^3 =2apq`

Options

A. `(2pq)/(a)`
B. 2apq
C. `2ap^(2)q^(2)`
D. none of these

Explanation

To solve the problem, we need to show that if one arithmetic mean (AM) \( a \) and two geometric means (GMs) \( p \) and \( q \) are inserted between two given numbers \( x \) and \( y \), then the following equation holds:

\[ p^3 + q^3 = 2apq \]

### Step-by-Step Solution:

1. **Define the Given Numbers**:
Let the two given numbers be \( x \) and \( y \).

2. **Insert the Arithmetic Mean**:
Since \( a \) is the arithmetic mean of \( x \) and \( y \), we have:
\[ a = \frac{x + y}{2} \]

3. **Set Up the Arithmetic Mean Condition**:
From the definition of the arithmetic mean, we can express the relationship:
\[ a - x = y - a \]
This implies:
\[ 2a = x + y \]

4. **Insert the Geometric Means**:
The two geometric means \( p \) and \( q \) are inserted between \( x \) and \( y \). According to the property of geometric means:
- For the first three terms \( x, p, q \):
\[ p^2 = xq \]
- For the last three terms \( p, q, y \):
\[ q^2 = py \]

5. **Express \( x \) and \( y \) in Terms of \( p \) and \( q \)**:
From the equation \( p^2 = xq \), we can express \( x \) as:
\[ x = \frac{p^2}{q} \]
From the equation \( q^2 = py \), we can express \( y \) as:
\[ y = \frac{q^2}{p} \]

6. **Substitute \( x \) and \( y \) into the AM Equation**:
Substitute the expressions for \( x \) and \( y \) into the equation \( 2a = x + y \):
\[ 2a = \frac{p^2}{q} + \frac{q^2}{p} \]

7. **Find a Common Denominator**:
The common denominator for the right-hand side is \( pq \):
\[ 2a = \frac{p^3 + q^3}{pq} \]

8. **Multiply Both Sides by \( pq \)**:
To eliminate the fraction, multiply both sides by \( pq \):
\[ 2apq = p^3 + q^3 \]

9. **Conclusion**:
We have shown that:
\[ p^3 + q^3 = 2apq \]
Hence, the required relation is proved.