Key Concepts and Formulas

• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The $n$-th term is given by $a_n = a_1 r^{n-1}$.
• Sum of first $n$ terms of GP ($S_n$): $S_n = a_1 + a_2 + \ldots + a_n$.
• Arithmetic Mean: The arithmetic mean of two numbers $x$ and $y$ is $\frac{x+y}{2}$.

Step-by-Step Solution

Step 1: Formulate the condition relating terms in the GP

The problem states that each term of the geometric progression is the arithmetic mean of the next two terms. For any term $a_k$, this can be written as: $a_k = \frac{a_{k+1} + a_{k+2}}{2}$ Multiplying both sides by 2, we get: $2a_k = a_{k+1} + a_{k+2}$

Step 2: Determine the common ratio ($r$) of the GP

We are given $a_1 = \frac{1}{8}$ and $a_2 \neq a_1$. We can use the relationship derived in Step 1 for the first few terms of the GP. Let's consider $a_1, a_2,$ and $a_3$: $2a_1 = a_2 + a_3$ Now, we express $a_2$ and $a_3$ in terms of $a_1$ and the common ratio $r$: $a_2 = a_1 r$ $a_3 = a_1 r^2$ Substitute these into the equation from Step 1: $2a_1 = a_1 r + a_1 r^2$ Since $a_1 = \frac{1}{8} \neq 0$, we can divide the entire equation by $a_1$: $2 = r + r^2$ Rearranging this into a quadratic equation: $r^2 + r - 2 = 0$ Factoring the quadratic equation: $(r+2)(r-1) = 0$ This gives two possible values for the common ratio: $r = 1$ or $r = -2$.

The problem states that $a_2 \neq a_1$. If $r=1$, then $a_2 = a_1 \cdot 1 = a_1$, which contradicts the given condition. Therefore, the common ratio must be $r = -2$.

Step 3: Simplify the expression $S_{20} - S_{18}$

The sum of the first $n$ terms of a sequence is denoted by $S_n$. We are asked to find $S_{20} - S_{18}$. $S_{20} = a_1 + a_2 + \ldots + a_{18} + a_{19} + a_{20}$ $S_{18} = a_1 + a_2 + \ldots + a_{18}$ Subtracting $S_{18}$ from $S_{20}$: $S_{20} - S_{18} = (a_1 + a_2 + \ldots + a_{18} + a_{19} + a_{20}) - (a_1 + a_2 + \ldots + a_{18})$ $S_{20} - S_{18} = a_{19} + a_{20}$ This simplification is a key step, as it reduces the problem to finding only the 19th and 20th terms, rather than using the full sum formula.

Step 4: Calculate $a_{19} + a_{20}$

We know $a_1 = \frac{1}{8}$ and $r = -2$. The formula for the $n$-th term of a GP is $a_n = a_1 r^{n-1}$. $a_{19} = a_1 r^{19-1} = a_1 r^{18}$ $a_{20} = a_1 r^{20-1} = a_1 r^{19}$ Now, we sum these two terms: $a_{19} + a_{20} = a_1 r^{18} + a_1 r^{19}$ We can factor out the common term $a_1 r^{18}$: $a_{19} + a_{20} = a_1 r^{18} (1 + r)$ Substitute the values $a_1 = \frac{1}{8}$ and $r = -2$: $a_{19} + a_{20} = \left(\frac{1}{8}\right) (-2)^{18} (1 + (-2))$ Let's simplify the components:

• $\frac{1}{8} = 2^{-3}$
• $(-2)^{18} = 2^{18}$ (since the exponent is even)
• $1 + (-2) = -1$

Substitute these back into the expression: $a_{19} + a_{20} = (2^{-3}) (2^{18}) (-1)$ Using the exponent rule $x^m \cdot x^n = x^{m+n}$: $a_{19} + a_{20} = 2^{-3+18} (-1)$ $a_{19} + a_{20} = 2^{15} (-1)$ $a_{19} + a_{20} = -2^{15}$

Therefore, $S_{20} - S_{18} = -2^{15}$.

Common Mistakes & Tips

• Quadratic Equation Roots: When solving for the common ratio $r$, ensure you consider all possible roots. The condition $a_2 \neq a_1$ is critical for selecting the correct root.
• Exponent Rules with Negative Bases: Be careful when calculating powers of negative numbers. $(-x)^{\text{even exponent}} = x^{\text{even exponent}}$, and $(-x)^{\text{odd exponent}} = -x^{\text{odd exponent}}$.
• Simplifying $S_n - S_m$: Recognize that $S_n - S_m$ for $n > m$ always equals the sum of terms from $a_{m+1}$ to $a_n$. In this case, $S_{20} - S_{18} = a_{19} + a_{20}$.

Summary

The problem involved a geometric progression where each term was the arithmetic mean of the next two. This condition led to a quadratic equation for the common ratio, $r$. After determining $r=-2$ by using the constraint $a_2 \neq a_1$, we simplified the expression $S_{20} - S_{18}$ to $a_{19} + a_{20}$. Calculating these two terms using the GP formula $a_n = a_1 r^{n-1}$ and substituting the values of $a_1$ and $r$ yielded the final answer.

The final answer is $\boxed{-2^{15}}$.