Let's solve each part by using the properties of a geometric sequence, where the ratio between consecutive terms remains constant.

General Method:

For three consecutive terms $a$, $b$, and $c$, the condition is: $\frac{b}{a} = \frac{c}{b}$ This simplifies to: $b^2 = a \cdot c$

(a) $k, 3k, 54$

Using $b^2 = a \cdot c$: $(3k)^2 = k \cdot 54$ $9k^2 = 54k$ Divide by $k$ (assuming $k \neq 0$): $9k = 54$ $k = 6$

(b) $1000, 4k, k$

Using $b^2 = a \cdot c$: $(4k)^2 = 1000 \cdot k$ $16k^2 = 1000k$ Divide by $k$ (assuming $k \neq 0$): $16k = 1000$ $k = 62.5$

(c) $7, k, 28$

Using $b^2 = a \cdot c$: $k^2 = 7 \cdot 28$ $k^2 = 196$ $k = 14$

(d) $18, k, \frac{2}{9}$

Using $b^2 = a \cdot c$: $k^2 = 18 \cdot \frac{2}{9}$ $k^2 = 4$ k = 2 \quad \text{(or k = -2)}

(e) $k, 12, \frac{k}{9}$

Using $b^2 = a \cdot c$: $12^2 = k \cdot \frac{k}{9}$ $144 = \frac{k^2}{9}$ Multiply by 9: $1296 = k^2$ k = 36 \quad \text{(or k = -36)}

(f) $k, 20, \frac{25}{4}k$

Using $b^2 = a \cdot c$: $20^2 = k \cdot \frac{25}{4}k$ $400 = \frac{25k^2}{4}$ Multiply by 4: $1600 = 25k^2$ $k^2 = 64$ k = 8 \quad \text{(or k = -8)}

(g) $k, 3k, 20 - k$

Using $b^2 = a \cdot c$: $(3k)^2 = k \cdot (20 - k)$ $9k^2 = 20k - k^2$ $10k^2 - 20k = 0$ Factorize: $10k(k - 2) = 0$ k = 0 \quad \text{(not valid as k \neq 0)}, \quad k = 2

Let me know which part you'd like further clarified, or if you'd like all solutions detailed further.

Related Questions:

1. What are the general conditions for a geometric sequence?
2. Can a geometric sequence have negative terms?
3. How does the common ratio affect the terms of a geometric sequence?
4. What happens if $k = 0$ in any of these problems?
5. Can you determine $k$ if four terms of the sequence are given?

Tip:

Always simplify and check the sequence properties (like $b^2 = a \cdot c$) step by step to avoid errors!