If the arithmetic mean and geometric mean of the `p^(th) and q^(th)` terms of the sequence -16,8,-4,2,… satisfy the equation
`4x^2-9x+5=0` , then p+q is equal to `"______"`

Answer

To solve the problem, we need to find the values of \( p \) and \( q \) such that the arithmetic mean (AM) and geometric mean (GM) of the \( p^{th} \) and \( q^{th} \) terms of the sequence satisfy the equation \( 4x^2 - 9x + 5 = 0 \).

### Step 1: Identify the sequence
The given sequence is:
\[ -16, 8, -4, 2, \ldots \]
This sequence is a geometric progression (GP) where the first term \( a = -16 \) and the common ratio \( r = -\frac{1}{2} \).

### Step 2: Find the \( p^{th} \) and \( q^{th} \) terms
The \( n^{th} \) term of a GP can be calculated using the formula:
\[ T_n = a \cdot r^{n-1} \]
Thus, the \( p^{th} \) term is:
\[ T_p = -16 \left(-\frac{1}{2}\right)^{p-1} = -16 \cdot \left(-\frac{1}{2}\right)^{p-1} \]
And the \( q^{th} \) term is:
\[ T_q = -16 \left(-\frac{1}{2}\right)^{q-1} = -16 \cdot \left(-\frac{1}{2}\right)^{q-1} \]

### Step 3: Calculate the Arithmetic Mean (AM)
The arithmetic mean of \( T_p \) and \( T_q \) is given by:
\[ AM = \frac{T_p + T_q}{2} \]

### Step 4: Calculate the Geometric Mean (GM)
The geometric mean of \( T_p \) and \( T_q \) is given by:
\[ GM = \sqrt{T_p \cdot T_q} \]

### Step 5: Set up the equations
We know that the AM and GM are the roots of the equation \( 4x^2 - 9x + 5 = 0 \). We can find the roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = -9 \), and \( c = 5 \):
\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 4 \cdot 5}}{2 \cdot 4} \]
\[ x = \frac{9 \pm \sqrt{81 - 80}}{8} = \frac{9 \pm 1}{8} \]
Thus, the roots are:
\[ x_1 = \frac{10}{8} = \frac{5}{4}, \quad x_2 = \frac{8}{8} = 1 \]

### Step 6: Assign values to AM and GM
Let:
\[ AM = \frac{5}{4}, \quad GM = 1 \]

### Step 7: Relate AM and GM to \( T_p \) and \( T_q \)
From the definitions:
\[ \frac{T_p + T_q}{2} = \frac{5}{4} \implies T_p + T_q = \frac{5}{2} \]
\[ \sqrt{T_p \cdot T_q} = 1 \implies T_p \cdot T_q = 1 \]

### Step 8: Set up the equations
Let \( T_p = x \) and \( T_q = y \). Then we have:
1. \( x + y = \frac{5}{2} \)
2. \( xy = 1 \)

### Step 9: Solve the equations
From the first equation, we can express \( y \):
\[ y = \frac{5}{2} - x \]
Substituting into the second equation:
\[ x\left(\frac{5}{2} - x\right) = 1 \]
\[ \frac{5}{2}x - x^2 = 1 \implies x^2 - \frac{5}{2}x + 1 = 0 \]

### Step 10: Find \( p + q \)
Using the quadratic formula again:
\[ x = \frac{\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4 \cdot 1}}{2} \]
\[ x = \frac{\frac{5}{2} \pm \sqrt{\frac{25}{4} - 4}}{2} = \frac{\frac{5}{2} \pm \sqrt{\frac{9}{4}}}{2} \]
\[ x = \frac{\frac{5}{2} \pm \frac{3}{2}}{2} \]
Thus, the roots are:
\[ x_1 = 2, \quad x_2 = \frac{1}{2} \]

### Step 11: Find \( p + q \)
Now substituting back, we find \( p + q \):
\[ p + q = 10 \]

Thus, the final answer is:
\[ \boxed{10} \]