The first, second and seventh terms of an arithmetic progression (all the terms are distinct) are in geometric progression and the sum of these three terms is 93. Then, the fourth term of this geometric progression is

Options

A. 21
B. 31
C. 75
D. 375

Answer

To solve the problem, we need to find the fourth term of a geometric progression (GP) formed by the first, second, and seventh terms of an arithmetic progression (AP). Here are the steps to arrive at the solution:

### Step 1: Define the terms of the AP
Let the first term of the arithmetic progression be \( a \) and the common difference be \( d \). Then the terms can be expressed as:
- First term \( T_1 = a \)
- Second term \( T_2 = a + d \)
- Seventh term \( T_7 = a + 6d \)

### Step 2: Set up the GP condition
According to the problem, the first, second, and seventh terms of the AP are in geometric progression. For three numbers \( x, y, z \) to be in GP, the condition is:
\[ y^2 = x \cdot z \]
Applying this to our terms:
\[ (a + d)^2 = a \cdot (a + 6d) \]

### Step 3: Expand and simplify the equation
Expanding both sides gives:
\[ a^2 + 2ad + d^2 = a^2 + 6ad \]
Now, subtract \( a^2 \) from both sides:
\[ 2ad + d^2 = 6ad \]
Rearranging gives:
\[ d^2 - 4ad = 0 \]

### Step 4: Factor the equation
Factoring out \( d \) gives:
\[ d(d - 4a) = 0 \]
Since the terms are distinct, \( d \neq 0 \), thus:
\[ d - 4a = 0 \implies d = 4a \]

### Step 5: Use the sum condition
The sum of the three terms is given as 93:
\[ T_1 + T_2 + T_7 = 93 \]
Substituting the terms:
\[ a + (a + d) + (a + 6d) = 93 \]
This simplifies to:
\[ 3a + 7d = 93 \]

### Step 6: Substitute \( d \) in the sum equation
Now substitute \( d = 4a \) into the sum equation:
\[ 3a + 7(4a) = 93 \]
This simplifies to:
\[ 3a + 28a = 93 \implies 31a = 93 \]
Solving for \( a \):
\[ a = \frac{93}{31} = 3 \]

### Step 7: Calculate \( d \)
Now substitute \( a \) back to find \( d \):
\[ d = 4a = 4 \times 3 = 12 \]

### Step 8: Find the terms of the AP
Now we can find the terms:
- \( T_1 = a = 3 \)
- \( T_2 = a + d = 3 + 12 = 15 \)
- \( T_7 = a + 6d = 3 + 6 \times 12 = 3 + 72 = 75 \)

### Step 9: Find the common ratio of the GP
The terms \( 3, 15, 75 \) are in GP. The common ratio \( r \) is:
\[ r = \frac{T_2}{T_1} = \frac{15}{3} = 5 \]

### Step 10: Find the fourth term of the GP
The \( n \)-th term of a GP is given by:
\[ T_n = ar^{n-1} \]
For the fourth term:
\[ T_4 = 3 \cdot 5^{4-1} = 3 \cdot 5^3 = 3 \cdot 125 = 375 \]

### Final Answer
Thus, the fourth term of the geometric progression is \( \boxed{375} \).