Let `{A_n}` be a unique sequence of positive integers satisfying the following properties:
`A_1 = 1, A_2 = 2, A_4 = 12, and A_(n+1) . A_(n-1) = A_n^2 pm 1 `for n = 2,3,4… then , `A_7` is

Options

A. 60
B. 120
C. 149
D. 169

Answer

To find \( A_7 \) in the sequence defined by the properties given, we will follow these steps:

### Step 1: Use the given values to find \( A_3 \)
We know:
- \( A_1 = 1 \)
- \( A_2 = 2 \)
- \( A_4 = 12 \)

Using the recurrence relation for \( n = 2 \):
\[ A_{n+1} \cdot A_{n-1} = A_n^2 \pm 1 \]
Substituting \( n = 2 \):
\[ A_3 \cdot A_1 = A_2^2 \pm 1 \]
This simplifies to:
\[ A_3 \cdot 1 = 2^2 \pm 1 \]
\[ A_3 = 4 \pm 1 \]
Thus, \( A_3 \) can be either \( 5 \) or \( 3 \).

### Step 2: Use the values of \( A_3 \) to find \( A_5 \)
Now, we will use both possible values of \( A_3 \) to find \( A_5 \).

**Case 1:** If \( A_3 = 5 \):
Using the recurrence relation for \( n = 3 \):
\[ A_4 \cdot A_2 = A_3^2 \pm 1 \]
Substituting:
\[ 12 \cdot 2 = 5^2 \pm 1 \]
\[ 24 = 25 \pm 1 \]
This gives us:
\[ 24 = 26 \quad \text{(not valid)} \]
So, \( A_3 \) cannot be \( 5 \).

**Case 2:** If \( A_3 = 3 \):
Using the same recurrence relation:
\[ A_4 \cdot A_2 = A_3^2 \pm 1 \]
Substituting:
\[ 12 \cdot 2 = 3^2 \pm 1 \]
\[ 24 = 9 \pm 1 \]
This gives us:
\[ 24 = 10 \quad \text{(not valid)} \]
So, we need to check again.

### Step 3: Find \( A_5 \) using \( A_3 = 5 \)
Now, let's use \( A_3 = 5 \) again for \( A_5 \):
Using the recurrence relation for \( n = 4 \):
\[ A_5 \cdot A_3 = A_4^2 \pm 1 \]
Substituting:
\[ A_5 \cdot 5 = 12^2 \pm 1 \]
\[ A_5 \cdot 5 = 144 \pm 1 \]
This gives us two cases:
1. \( A_5 \cdot 5 = 145 \) → \( A_5 = 29 \)
2. \( A_5 \cdot 5 = 143 \) → \( A_5 = 28.6 \) (not valid since \( A_5 \) must be an integer)

So, \( A_5 = 29 \).

### Step 4: Find \( A_6 \)
Now, we can find \( A_6 \) using \( A_5 = 29 \):
Using the recurrence relation for \( n = 5 \):
\[ A_6 \cdot A_4 = A_5^2 \pm 1 \]
Substituting:
\[ A_6 \cdot 12 = 29^2 \pm 1 \]
Calculating \( 29^2 = 841 \):
\[ A_6 \cdot 12 = 841 \pm 1 \]
This gives us two cases:
1. \( A_6 \cdot 12 = 842 \) → \( A_6 = 70.1667 \) (not valid)
2. \( A_6 \cdot 12 = 840 \) → \( A_6 = 70 \)

### Step 5: Find \( A_7 \)
Finally, we find \( A_7 \) using \( A_6 \):
Using the recurrence relation for \( n = 6 \):
\[ A_7 \cdot A_5 = A_6^2 \pm 1 \]
Substituting:
\[ A_7 \cdot 29 = 70^2 \pm 1 \]
Calculating \( 70^2 = 4900 \):
\[ A_7 \cdot 29 = 4900 \pm 1 \]
This gives us two cases:
1. \( A_7 \cdot 29 = 4901 \) → \( A_7 = 169 \)
2. \( A_7 \cdot 29 = 4899 \) → \( A_7 = 168.2414 \) (not valid)

Thus, we conclude:
\[ A_7 = 169 \]

### Final Answer:
The value of \( A_7 \) is \( 169 \).