a, b, c are the first three terms of a geometric series. If the harmonic mean of a and b is 12 and that of b and c is 36, then the first five terms of the series are………………..

Answer

To solve the problem, we need to find the first five terms of a geometric series given the harmonic means of the first two terms and the last two terms. Let's denote the first three terms of the geometric series as \( a, b, c \).

### Step 1: Understanding the Harmonic Mean
The harmonic mean (HM) of two numbers \( x \) and \( y \) is given by the formula:
\[ HM = \frac{2xy}{x + y} \]

### Step 2: Setting Up the Equations
From the problem, we know:
1. The harmonic mean of \( a \) and \( b \) is 12:
\[ \frac{2ab}{a + b} = 12 \]
Rearranging gives:
\[ 2ab = 12(a + b) \quad \Rightarrow \quad 2ab - 12a - 12b = 0 \quad \Rightarrow \quad ab - 6a - 6b = 0 \quad \Rightarrow \quad ab = 6a + 6b \quad (1) \]

2. The harmonic mean of \( b \) and \( c \) is 36:
\[ \frac{2bc}{b + c} = 36 \]
Rearranging gives:
\[ 2bc = 36(b + c) \quad \Rightarrow \quad 2bc - 36b - 36c = 0 \quad \Rightarrow \quad bc - 18b - 18c = 0 \quad \Rightarrow \quad bc = 18b + 18c \quad (2) \]

### Step 3: Expressing \( a \) and \( c \) in Terms of \( b \)
Since \( a, b, c \) are in a geometric progression, we can express \( a \) and \( c \) in terms of \( b \) and the common ratio \( r \):
- Let \( a = \frac{b}{r} \)
- Let \( c = br \)

### Step 4: Substituting into the Equations
Substituting \( a \) and \( c \) into equations (1) and (2):

1. From equation (1):
\[ \left(\frac{b}{r}\right)b = 6\left(\frac{b}{r}\right) + 6b \]
Simplifying gives:
\[ \frac{b^2}{r} = \frac{6b}{r} + 6b \]
Multiplying through by \( r \):
\[ b^2 = 6b + 6br \quad \Rightarrow \quad b^2 - 6b - 6br = 0 \quad (3) \]

2. From equation (2):
\[ b(br) = 18b + 18(br) \]
Simplifying gives:
\[ b^2r = 18b + 18br \]
Rearranging gives:
\[ b^2r - 18br - 18b = 0 \quad (4) \]

### Step 5: Solving the Equations
Now we have two equations (3) and (4) in terms of \( b \) and \( r \). We can solve these equations simultaneously.

From (3):
\[ b(b - 6 - 6r) = 0 \]
This gives us two cases: \( b = 0 \) (not valid) or \( b - 6 - 6r = 0 \) which implies:
\[ b = 6 + 6r \]

From (4):
\[ b(b r - 18 - 18) = 0 \]
This gives us \( b = 0 \) (not valid) or \( br = 18 + 18 \) which implies:
\[ br = 18 + 18 = 36 \]

### Step 6: Finding Values
Substituting \( b = 6 + 6r \) into \( br = 36 \):
\[ (6 + 6r)r = 36 \]
Expanding gives:
\[ 6r + 6r^2 = 36 \quad \Rightarrow \quad r^2 + r - 6 = 0 \]
Factoring gives:
\[ (r - 2)(r + 3) = 0 \quad \Rightarrow \quad r = 2 \text{ (valid) or } r = -3 \text{ (not valid)} \]
Thus, \( r = 2 \).

### Step 7: Finding \( b \)
Substituting \( r = 2 \) back into \( b = 6 + 6r \):
\[ b = 6 + 6(2) = 6 + 12 = 18 \]

### Step 8: Finding \( a \) and \( c \)
Now we can find \( a \) and \( c \):
\[ a = \frac{b}{r} = \frac{18}{2} = 9 \]
\[ c = br = 18 \cdot 2 = 36 \]

### Step 9: First Five Terms of the Series
The first three terms of the geometric series are \( a, b, c \):
- \( a = 9 \)
- \( b = 18 \)
- \( c = 36 \)

The first five terms of the geometric series are:
1. \( 9 \)
2. \( 18 \)
3. \( 36 \)
4. \( 72 \) (which is \( c \cdot r = 36 \cdot 2 \))
5. \( 144 \) (which is \( 72 \cdot 2 \))

### Final Answer
The first five terms of the series are:
\[ 9, 18, 36, 72, 144 \]