lf n! ,`3xxn!` and (n+1)! are in G P, then n!, 5 x n! and (n+1)! are in

Options

A. i) AP
B. ii) GP
C. iii) HP
D. iv) AGP

Answer

To solve the problem step by step, we will analyze the given information and derive the necessary conclusions.

### Step 1: Understand the Given Information
We are given that \( n! \), \( 3 \times n! \), and \( (n+1)! \) are in a geometric progression (G.P.).

### Step 2: Set Up the G.P. Condition
For three terms \( a \), \( b \), and \( c \) to be in G.P., the condition is:
\[ b^2 = a \cdot c \]
In our case:
- \( a = n! \)
- \( b = 3 \times n! \)
- \( c = (n+1)! \)

Thus, we can write:
\[ (3 \times n!)^2 = n! \cdot (n+1)! \]

### Step 3: Simplify the Equation
Substituting the factorials, we have:
\[ 9 \times (n!)^2 = n! \cdot (n+1) \cdot n! \]
This simplifies to:
\[ 9 \times (n!)^2 = n! \cdot n! \cdot (n+1) \]
Dividing both sides by \( (n!)^2 \) (assuming \( n! \neq 0 \)):
\[ 9 = n + 1 \]

### Step 4: Solve for \( n \)
From the equation \( 9 = n + 1 \), we can solve for \( n \):
\[ n = 9 - 1 = 8 \]

### Step 5: Substitute \( n \) into the New Terms
Now we need to check if \( n! \), \( 5 \times n! \), and \( (n+1)! \) are in a certain progression. We substitute \( n = 8 \):
- \( n! = 8! = 40320 \)
- \( 5 \times n! = 5 \times 8! = 5 \times 40320 = 201600 \)
- \( (n+1)! = 9! = 9 \times 8! = 9 \times 40320 = 362880 \)

### Step 6: Check the Progression
We denote:
- \( a_1 = 8! = 40320 \)
- \( a_2 = 5 \times 8! = 201600 \)
- \( a_3 = 9! = 362880 \)

To check if these values are in arithmetic progression (A.P.), we need to verify:
\[ a_2 - a_1 = a_3 - a_2 \]

Calculating:
- \( a_2 - a_1 = 201600 - 40320 = 161280 \)
- \( a_3 - a_2 = 362880 - 201600 = 161280 \)

Since both differences are equal, we conclude that:
\[ a_1, a_2, a_3 \text{ are in A.P.} \]

### Final Answer
Thus, \( n! \), \( 5 \times n! \), and \( (n+1)! \) are in **A.P.**

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