If the arithmetic mean and geometric mean of the pth and qth terms of the sequence -16, 8, -4, 2, ..

Q: Wait, why do we reject the negative root here?

Actually, we don't reject roots based on sign, but based on the $AM \ge GM$ property. Since $AM$ must be $\ge GM$ for positive terms, and $\frac{5}{4} > 1$, the assignment is fixed. We are not rejecting a root; we are assigning them to their correct mathematical roles.

Q: Why can't we just use energy conservation?

This is a sequence problem, not physics! But the spirit is the same: conservation of logic. We use the properties of G.P. terms to conserve the relationship between the variables $p$ and $q$ throughout the algebraic manipulation.

Q: How do I know if $p+q-2$ is even?

Great question! We equate the exponents *after* ensuring the bases match. Since $(-\frac{1}{2})^8$ is positive, and our equation resulted in a positive value ($1/256$), the exponent must be even for the equality to hold. It is a necessary condition for the solution to exist.

Question

If the arithmetic mean and geometric mean of the $p^{th}$ and $q^{th}$ terms of the sequence $-16, 8, -4, 2, \dots$ satisfy the equation $4x^2 - 9x + 5 = 0$, then $p + q$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Sequence

Imagine you are standing on the edge of a mathematical landscape, looking at the sequence $-16, 8, -4, 2, \dots$. At first glance, it might seem like a chaotic oscillation, but the ratio between consecutive terms is constant.

We calculate the common ratio $r$ as:
$$\frac{8}{-16} = -\frac{1}{2}, \quad \frac{-4}{8} = -\frac{1}{2}$$

This is the heartbeat of a **Geometric Progression (G.P.)**. We identify our first term $a = -16$ and our common ratio $r = -1/2$.

### The Quadratic Gatekeeper

Now, we encounter a quadratic equation: $4x^2 - 9x + 5 = 0$. This equation acts as a gatekeeper, holding the keys to our Arithmetic Mean ($AM$) and Geometric Mean ($GM$).

By splitting the middle term, we factorize it into $(4x - 5)(x - 1) = 0$. This yields two roots: $x = 1$ and $x = 5/4$.

We invoke the wisdom of the **$AM \ge GM$ inequality**. Since $5/4 > 1$, the $AM$ must be $5/4$ and the $GM$ must be $1$. This is a crucial realization—the $GM$ is $1$, which implies the product of our two terms, $t_p$ and $t_q$, must be $1^2 = 1$.

### The Exponential Journey

Let us utilize the general term of our G.P.: $t_n = a r^{n-1}$. Substituting our known values, we express the terms as:
$$t_p = -16\left(-\frac{1}{2}\right)^{p-1}, \quad t_q = -16\left(-\frac{1}{2}\right)^{q-1}$$

When we multiply these to satisfy the $GM$ condition, we obtain:
$$\left[-16\left(-\frac{1}{2}\right)^{p-1}\right] \cdot \left[-16\left(-\frac{1}{2}\right)^{q-1}\right] = 1$$

Multiplying the constants $-16 \cdot -16$ gives us $256$. Adding the exponents of the common ratio, we get $(p-1) + (q-1) = p+q-2$. Our equation simplifies to:
$$256 \cdot \left(-\frac{1}{2}\right)^{p+q-2} = 1$$

### The Final Symmetry

Dividing both sides by $256$, we arrive at:
$$\left(-\frac{1}{2}\right)^{p+q-2} = \frac{1}{256}$$

We know that $256 = 2^8$, so $1/256 = (1/2)^8$. Because $8$ is an even number, $(1/2)^8$ is identical to $(-1/2)^8$.

Now, our bases match perfectly. We equate the exponents:
$$p + q - 2 = 8$$

Solving this gives us the final result:
**$p + q = 10$**

Animated Solution for Mathematics - Sequence and Series: If the arithmetic mean and geometric mean of the $p^{t h}$ and $q^{t h}$ terms of the sequence $- 16, 8, - 4, 2, \dots$ satisfy the equation $4 x^{2} - 9 x + 5 = 0$ , then $p + q$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Sequence

Solve the Quadratic Equation

Identify $A M$ and $GM$

Set up the $GM$ Equation

Substitute Terms into Equation

Simplify the Constants

Isolate the Exponential Term

Match the Bases

Solve for $p + q$

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Sequence

The Quadratic Gatekeeper

The Exponential Journey

The Final Symmetry

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