The image contains problems related to geometric sequences. Let's go through each question step-by-step.

1. Determine if the sequence is geometric. If it is, find the common ratio.

The sequences are:

1. $1, -3, 9, -27, \ldots$
2. $5, 15, 45, 135, \ldots$
3. $\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots$
4. $2, 5, 8, 11, \ldots$

Solution:

• (a)

  • $r = \frac{-3}{1} = -3, \; r = \frac{9}{-3} = -3$ (Common ratio = -3)
  • Geometric sequence with common ratio $r = -3$.

• (b)

  • $r = \frac{15}{5} = 3, \; r = \frac{45}{15} = 3$ (Common ratio = 3)
  • Geometric sequence with common ratio $r = 3$.

• (c)

  • $r = \frac{1/6}{1/3} = \frac{1}{2}, \; r = \frac{1/12}{1/6} = \frac{1}{2}$ (Common ratio = 1/2)
  • Geometric sequence with common ratio $r = \frac{1}{2}$.

• (d)

  • Difference between terms: $5 - 2 = 3, \; 8 - 5 = 3, \; 11 - 8 = 3$ (Common difference = 3)
  • Not a geometric sequence.

2. Given the explicit formula of a geometric sequence, find the first five terms.

Explicit formulas:

1. $a_n = 5(-2)^{n-1}$
2. $a_n = \frac{2}{3} (4)^{n-1}$

Solution:

• (a)
  $a_1 = 5(-2)^0 = 5, \; a_2 = 5(-2)^1 = -10, \; a_3 = 5(-2)^2 = 20, \; a_4 = 5(-2)^3 = -40, \; a_5 = 5(-2)^4 = 80$
  First five terms: $5, -10, 20, -40, 80$.

• (b)
  $a_1 = \frac{2}{3} (4)^0 = \frac{2}{3}, \; a_2 = \frac{2}{3} (4)^1 = \frac{8}{3}, \; a_3 = \frac{2}{3} (4)^2 = \frac{32}{3}, \; a_4 = \frac{2}{3} (4)^3 = \frac{128}{3}, \; a_5 = \frac{2}{3} (4)^4 = \frac{512}{3}$
  First five terms: $\frac{2}{3}, \frac{8}{3}, \frac{32}{3}, \frac{128}{3}, \frac{512}{3}$.

3. Find the indicated term of the geometric sequence.

Given sequences:

1. $a = 4, r = 3, n = 7$
2. $a = 5, r = -2, n = 8$

Solution:

• (a) $a_n = ar^{n-1} = 4 \times 3^{6} = 4 \times 729 = 2916$
• (b) $a_n = ar^{n-1} = 5 \times (-2)^{7} = 5 \times (-128) = -640$

4. Find the sum, $S_n$, for the first $n$ terms of the given geometric sequence.

Given sequences:

1. $a = 2, r = 3, n = 5$
2. $a = 6, r = \frac{1}{2}, n = 6$

Solution:

• (a)
  $S_n = a \frac{r^n - 1}{r - 1} = 2 \frac{3^5 - 1}{2} = 2 \frac{243 - 1}{2} = 2 \times 121 = 242$

• (b)
  $S_n = a \frac{1 - r^n}{1 - r} = 6 \frac{1 - (\frac{1}{2})^6}{\frac{1}{2}} = 6 \frac{1 - \frac{1}{64}}{\frac{1}{2}} = 6 \times \frac{\frac{63}{64}}{\frac{1}{2}} = 6 \times \frac{63}{32} = \frac{378}{32} = 11.8125$

5. Find the indicated geometric sum.

Given:

1. $\sum_{k=1}^{8} 2 \cdot (3)^{k-1}$
2. $\sum_{k=1}^{5} 5 \cdot \left(\frac{1}{2}\right)^{k-1}$

Solution:

• (a)
  $S_8 = 2 \frac{3^8 - 1}{3 - 1} = 2 \frac{6561 - 1}{2} = 2 \times 3280 = 6560$

• (b)
  $S_5 = 5 \frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}} = 5 \frac{1 - \frac{1}{32}}{\frac{1}{2}} = 5 \times \frac{\frac{31}{32}}{\frac{1}{2}} = 5 \times \frac{31}{16} = \frac{155}{16} = 9.6875$

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Relative Questions

1. How do you determine if a sequence is geometric?
2. What is the formula for the $n$-th term of a geometric sequence?
3. How can you find the sum of an infinite geometric series?
4. What is the significance of the common ratio in a geometric sequence?
5. How does the explicit formula of a geometric sequence differ from its recursive formula?

Tip

When working with geometric sequences, always check the common ratio by dividing consecutive terms to confirm the sequence's geometric nature.