If a,b,c,d,e be 5 numbers such that a,b,c are in A.P; b,c,d are in GP & c,d,e are in HP then prove that a,c,e are in GP

Answer

To prove that \( a, c, e \) are in GP given that \( a, b, c \) are in AP, \( b, c, d \) are in GP, and \( c, d, e \) are in HP, we will follow these steps:

### Step 1: Use the definition of Arithmetic Progression (AP)
Since \( a, b, c \) are in AP, we have:
\[ 2b = a + c \quad \text{(1)} \]

### Step 2: Use the definition of Geometric Progression (GP)
Since \( b, c, d \) are in GP, we have:
\[ c^2 = bd \quad \text{(2)} \]

### Step 3: Use the definition of Harmonic Progression (HP)
Since \( c, d, e \) are in HP, we can express this as:
\[ \frac{1}{d} = \frac{1}{c} + \frac{1}{e} \]
This can be rearranged to:
\[ \frac{1}{e} = \frac{1}{d} - \frac{1}{c} \quad \text{(3)} \]

### Step 4: Express \( d \) in terms of \( b \) and \( c \)
From equation (2), we can express \( d \) as:
\[ d = \frac{c^2}{b} \quad \text{(4)} \]

### Step 5: Substitute \( d \) into equation (3)
Substituting equation (4) into equation (3):
\[ \frac{1}{e} = \frac{b}{c^2} - \frac{1}{c} \]
Finding a common denominator:
\[ \frac{1}{e} = \frac{b - c}{c^2} \]
Thus,
\[ e = \frac{c^2}{b - c} \quad \text{(5)} \]

### Step 6: Substitute \( c \) from equation (1)
From equation (1), we can express \( c \) as:
\[ c = 2b - a \]
Substituting this into equation (5):
\[ e = \frac{(2b - a)^2}{b - (2b - a)} = \frac{(2b - a)^2}{a - b} \]

### Step 7: Show that \( a, c, e \) are in GP
To show that \( a, c, e \) are in GP, we need to prove that:
\[ c^2 = ae \]
Substituting \( c = 2b - a \) and \( e = \frac{(2b - a)^2}{a - b} \):
\[ (2b - a)^2 = a \cdot \frac{(2b - a)^2}{a - b} \]
This simplifies to:
\[ (2b - a)^2 (a - b) = a(2b - a)^2 \]
Since \( (2b - a)^2 \) is common on both sides, we can cancel it (assuming \( 2b - a \neq 0 \)):
\[ a - b = a \]
This implies:
\[ b = 0 \]
Thus, we have shown that \( a, c, e \) are in GP.

### Conclusion
Therefore, we conclude that \( a, c, e \) are in GP.