a, b, c are the first three terms of geometric series. If the H.M. of a and b is 12 and that of b and c is 36 then which of the following hold(s) good?

Options

A. 9, 16, 27, 41, 58
B. 8, 24, 72, 216, 648
C. 4, 22, 38, 46, 57
D. 2, 6, 18, 54, 162

Answer

To solve the problem step by step, we will use the properties of geometric progression (GP) and harmonic mean (H.M.).

### Step 1: Understand the terms in the GP
Let the first three terms of the geometric progression be:
- \( a = A \)
- \( b = AR \)
- \( c = AR^2 \)

### Step 2: Use the given H.M. of \( a \) and \( b \)
The harmonic mean (H.M.) of two numbers \( x \) and \( y \) is given by the formula:
\[ H.M. = \frac{2xy}{x + y} \]
For \( a \) and \( b \):
\[ H.M. = \frac{2 \cdot a \cdot b}{a + b} = 12 \]
Substituting \( a \) and \( b \):
\[ \frac{2 \cdot A \cdot AR}{A + AR} = 12 \]
This simplifies to:
\[ \frac{2A^2R}{A(1 + R)} = 12 \]
Cancelling \( A \) from the numerator and denominator (assuming \( A \neq 0 \)):
\[ \frac{2AR}{1 + R} = 12 \]
Multiplying both sides by \( 1 + R \):
\[ 2AR = 12(1 + R) \]
This leads to:
\[ 2AR = 12 + 12R \]
Rearranging gives:
\[ 2AR - 12R = 12 \]
Factoring out \( R \):
\[ R(2A - 12) = 12 \]
Thus, we have:
\[ R = \frac{12}{2A - 12} \tag{1} \]

### Step 3: Use the given H.M. of \( b \) and \( c \)
Now, for \( b \) and \( c \):
\[ H.M. = \frac{2 \cdot b \cdot c}{b + c} = 36 \]
Substituting \( b \) and \( c \):
\[ \frac{2 \cdot AR \cdot AR^2}{AR + AR^2} = 36 \]
This simplifies to:
\[ \frac{2A^2R^3}{AR(1 + R)} = 36 \]
Cancelling \( AR \) (assuming \( AR \neq 0 \)):
\[ \frac{2AR^2}{1 + R} = 36 \]
Multiplying both sides by \( 1 + R \):
\[ 2AR^2 = 36(1 + R) \]
This leads to:
\[ 2AR^2 = 36 + 36R \]
Rearranging gives:
\[ 2AR^2 - 36R = 36 \]
Factoring out \( R \):
\[ R(2AR - 36) = 36 \]
Thus, we have:
\[ 2AR - 36 = \frac{36}{R} \tag{2} \]

### Step 4: Solve the equations
Now we have two equations (1) and (2). We can substitute \( R \) from equation (1) into equation (2).

From equation (1):
\[ R = \frac{12}{2A - 12} \]
Substituting into equation (2):
\[ 2A\left(\frac{12}{2A - 12}\right) - 36 = \frac{36}{\frac{12}{2A - 12}} \]
This simplifies to:
\[ \frac{24A}{2A - 12} - 36 = \frac{36(2A - 12)}{12} \]
Cross-multiplying and simplifying will lead to the values of \( A \) and \( R \).

### Step 5: Find the values of \( A \) and \( R \)
After solving the equations, we find:
- \( R = 3 \)
- \( A = 8 \)

### Step 6: Find the first five terms of the GP
The first five terms of the GP are:
- \( A = 8 \)
- \( AR = 8 \cdot 3 = 24 \)
- \( AR^2 = 8 \cdot 3^2 = 72 \)
- \( AR^3 = 8 \cdot 3^3 = 216 \)
- \( AR^4 = 8 \cdot 3^4 = 648 \)

Thus, the first five terms are \( 8, 24, 72, 216, 648 \).

### Conclusion
The correct option is \( B \) with the first five terms being \( 8, 24, 72, 216, 648 \).

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