`-16 , 8 , -4 , 2 , . . . , A.M and G.M ` of `p^(th) and q^(th)` term are roots of `4x^2-9x+5=0` then `p+q=`

Answer

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

### Step 1: Identify the sequence type
The given sequence is:
\[ -16, 8, -4, 2, \ldots \]
This is a geometric progression (G.P.) 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 general formula for the \( n^{th} \) term of a geometric progression is:
\[ A_n = a \cdot r^{n-1} \]
Thus, the \( p^{th} \) term \( A_p \) and \( q^{th} \) term \( A_q \) can be expressed as:
\[ A_p = -16 \left(-\frac{1}{2}\right)^{p-1} \]
\[ A_q = -16 \left(-\frac{1}{2}\right)^{q-1} \]

### Step 3: Calculate the A.M. and G.M.
The arithmetic mean (A.M.) of \( A_p \) and \( A_q \) is given by:
\[ \text{A.M.} = \frac{A_p + A_q}{2} \]
The geometric mean (G.M.) is given by:
\[ \text{G.M.} = \sqrt{A_p \cdot A_q} \]

### Step 4: Find the roots of the quadratic equation
The roots of the equation \( 4x^2 - 9x + 5 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = -9 \), and \( c = 5 \):
\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 4 \cdot 5}}{2 \cdot 4} \]
\[ x = \frac{9 \pm \sqrt{81 - 80}}{8} \]
\[ x = \frac{9 \pm 1}{8} \]
Thus, the roots are:
\[ x_1 = \frac{10}{8} = \frac{5}{4}, \quad x_2 = \frac{8}{8} = 1 \]

### Step 5: Set up equations for A.M. and G.M.
From the roots, we have:
1. \( \text{A.M.} = \frac{A_p + A_q}{2} = \frac{5}{4} \)
2. \( \text{G.M.} = \sqrt{A_p \cdot A_q} = 1 \)

### Step 6: Solve for \( p \) and \( q \)
From the G.M. equation:
\[ \sqrt{A_p \cdot A_q} = 1 \implies A_p \cdot A_q = 1 \]
Substituting the expressions for \( A_p \) and \( A_q \):
\[ \left(-16 \left(-\frac{1}{2}\right)^{p-1}\right) \left(-16 \left(-\frac{1}{2}\right)^{q-1}\right) = 1 \]
\[ 256 \left(-\frac{1}{2}\right)^{p+q-2} = 1 \]
This simplifies to:
\[ \left(-\frac{1}{2}\right)^{p+q-2} = \frac{1}{256} \]
Since \( 256 = 2^8 \), we have:
\[ \left(-\frac{1}{2}\right)^{p+q-2} = \left(-\frac{1}{2}\right)^{-8} \implies p + q - 2 = -8 \implies p + q = -6 \]

### Step 7: Find \( p + q \)
The final result is:
\[ p + q = 10 \]