If `a_n = (a_(n-1) xx a_(n-2))/(2), a_5 = -6 and a_6 = -18`, what is the value of `a_3` ?

Answer

To find the value of \( a_3 \) given the recurrence relation \( a_n = \frac{a_{n+1} \cdot a_{n-2}}{2} \), and the values \( a_5 = -6 \) and \( a_6 = -18 \), we will follow these steps:

### Step 1: Write down the recurrence relation
The recurrence relation is given as:
\[ a_n = \frac{a_{n+1} \cdot a_{n-2}}{2} \]

### Step 2: Substitute \( n = 5 \) into the relation
We substitute \( n = 5 \) into the recurrence relation:
\[ a_5 = \frac{a_6 \cdot a_3}{2} \]

### Step 3: Substitute known values
We know that \( a_5 = -6 \) and \( a_6 = -18 \). Substituting these values into the equation gives:
\[ -6 = \frac{-18 \cdot a_3}{2} \]

### Step 4: Simplify the equation
To eliminate the fraction, multiply both sides by 2:
\[ -12 = -18 \cdot a_3 \]

### Step 5: Solve for \( a_3 \)
Now, divide both sides by -18:
\[ a_3 = \frac{-12}{-18} = \frac{12}{18} = \frac{2}{3} \]

### Final Answer
Thus, the value of \( a_3 \) is:
\[ \boxed{\frac{2}{3}} \]