Find the G.P. whose 2nd term is 12 and 6th term is 27 times the 3rd term.

Explanation

To find the geometric progression (G.P.) whose second term is 12 and the sixth term is 27 times the third term, we can follow these steps:

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)
- Sixth term: \( ar^5 \)

### Step 2: Set up the equations based on the given information
From the problem, we know:
1. The second term \( ar = 12 \)
2. The sixth term \( ar^5 = 27 \times \text{(third term)} = 27 \times ar^2 \)

### Step 3: Substitute the known values
From the second term:
\[ ar = 12 \]
From the sixth term:
\[ ar^5 = 27 \times ar^2 \]

### Step 4: Simplify the sixth term equation
Substituting \( ar^5 \) in the sixth term equation:
\[ ar^5 = 27 \times ar^2 \]
Dividing both sides by \( ar^2 \) (assuming \( a \neq 0 \) and \( r \neq 0 \)):
\[ \frac{ar^5}{ar^2} = 27 \]
This simplifies to:
\[ r^3 = 27 \]

### Step 5: Solve for \( r \)
Taking the cube root of both sides:
\[ r = 3 \]

### Step 6: Substitute \( r \) back to find \( a \)
Now, substitute \( r \) back into the equation for the second term:
\[ ar = 12 \]
\[ a \cdot 3 = 12 \]
Dividing both sides by 3:
\[ a = 4 \]

### Step 7: Write the G.P.
Now that we have \( a \) and \( r \):
- First term: \( a = 4 \)
- Second term: \( ar = 4 \cdot 3 = 12 \)
- Third term: \( ar^2 = 4 \cdot 3^2 = 4 \cdot 9 = 36 \)
- Fourth term: \( ar^3 = 4 \cdot 3^3 = 4 \cdot 27 = 108 \)
- Fifth term: \( ar^4 = 4 \cdot 3^4 = 4 \cdot 81 = 324 \)
- Sixth term: \( ar^5 = 4 \cdot 3^5 = 4 \cdot 243 = 972 \)

Thus, the G.P. is:
\[ 4, 12, 36, 108, 324, 972, \ldots \]

### Summary of the G.P.
The G.P. is \( 4, 12, 36, 108, 324, 972 \).

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