If `a,b,c` are in `H.P`, `b,c,d` are in `G.P` and `c,d,e` are in `A.P.` , then the value of `e` is (a) `(ab^(2))/((2a-b)^(2))` (b) `(a^(2)b)/((2a-b)^(2))` (c) `(a^(2)b^(2))/((2a-b)^(2))` (d) None of these

Options

A. `(ab^(2))/((2a-b)^(2))`
B. `(a^(2)b)/((2a-b)^(2))`
C. `(a^(2)b^(2))/((2a-b)^(2))`
D. None of these

Explanation

To solve the problem step by step, we will use the properties of Harmonic Progression (H.P.), Geometric Progression (G.P.), and Arithmetic Progression (A.P.) as given in the question.

### Step 1: Understanding H.P.
Given that \( a, b, c \) are in H.P., we know that the reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P.

Using the property of A.P., we have:
\[ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \]
This can be rearranged to:
\[ \frac{2}{b} = \frac{1}{c} + \frac{1}{a} \]
Multiplying through by \( abc \) gives:
\[ 2ac = ab + bc \]
Rearranging gives:
\[ \frac{1}{c} = \frac{2}{b} - \frac{1}{a} \]
Taking the common denominator:
\[ \frac{1}{c} = \frac{2a - b}{ab} \]
Thus,
\[ c = \frac{ab}{2a - b} \]

### Step 2: Understanding G.P.
Next, we know that \( b, c, d \) are in G.P. This means:
\[ c^2 = bd \]
Substituting the value of \( c \):
\[ \left(\frac{ab}{2a - b}\right)^2 = b \cdot d \]
From this, we can express \( d \):
\[ d = \frac{c^2}{b} = \frac{\left(\frac{ab}{2a - b}\right)^2}{b} = \frac{a^2b}{(2a - b)^2} \]

### Step 3: Understanding A.P.
Now, we know that \( c, d, e \) are in A.P. Therefore:
\[ 2d = c + e \]
Rearranging gives:
\[ e = 2d - c \]
Substituting the values of \( d \) and \( c \):
\[ e = 2\left(\frac{a^2b}{(2a - b)^2}\right) - \frac{ab}{2a - b} \]

### Step 4: Simplifying the Expression for \( e \)
Now we need to simplify this expression:
\[ e = \frac{2a^2b}{(2a - b)^2} - \frac{ab(2a - b)}{(2a - b)^2} \]
Combining the fractions:
\[ e = \frac{2a^2b - ab(2a - b)}{(2a - b)^2} \]
Expanding the numerator:
\[ e = \frac{2a^2b - 2a^2b + ab^2}{(2a - b)^2} \]
This simplifies to:
\[ e = \frac{ab^2}{(2a - b)^2} \]

### Final Answer
Thus, the value of \( e \) is:
\[ \boxed{\frac{ab^2}{(2a - b)^2}} \]