Classwork 1. $$ k+1 ; k-1 ; 2 k-5 ; \ldots $$ are the first three terms of a geometric sequence, where $$ k

Question

Classwork 1. $$ k+1 ; k-1 ; 2 k-5 ; \ldots $$ are the first three terms of a geometric sequence, where $$ k< $$ (a) Calculate the value of $$ k $$. (b) Determine the sequence. (c) Determine the general term of the sequence. (d) Find the 10 th term. (e) Which term of the sequence is -59049 ? 2. Determine an expression for the $$ \mathrm{n}^{	ext {th }} $$ term for a geometric sequence whose $$ 3^{	ext {th }} $$ term is -20 an the $$ 6^{	ext {th }} $$ term is 160 .

UpStudy AI · Accepted Answer

Let's tackle Problem 1 step by step.

### 1. Geometric Sequence: $$ k+1 ; k-1 ; 2k-5 ; \ldots $$

**(a) Calculate the value of $$ k $$.**

In a geometric sequence, the ratio between consecutive terms is constant. Let's denote the common ratio by $$ r $$.

Given the first three terms:
- First term: $$ a_1 = k + 1 $$
- Second term: $$ a_2 = k - 1 $$
- Third term: $$ a_3 = 2k - 5 $$

The common ratio $$ r $$ can be expressed as:
$$
r = \frac{a_2}{a_1} = \frac{k - 1}{k + 1}
$$
$$
r = \frac{a_3}{a_2} = \frac{2k - 5}{k - 1}
$$

Since both expressions equal $$ r $$, set them equal to each other:
$$
\frac{k - 1}{k + 1} = \frac{2k - 5}{k - 1}
$$

**Cross-multiply to solve for $$ k $$:**
$$
(k - 1)^2 = (2k - 5)(k + 1)
$$
$$
k^2 - 2k + 1 = 2k^2 - 4k - 5
$$

**Rearrange terms:**
$$
0 = 2k^2 - 4k - 5 - k^2 + 2k - 1
$$
$$
0 = k^2 - 2k - 6
$$

**Solve the quadratic equation:**
$$
k^2 - 2k - 6 = 0
$$
$$
k = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}
$$

**Possible values of $$ k $$:**
$$
k = 1 + \sqrt{7} \quad \text{or} \quad k = 1 - \sqrt{7}
$$

**Considerations:**
- If $$ k = 1 + \sqrt{7} $$, all terms $$ k+1 $$, $$ k-1 $$, and $$ 2k-5 $$ are positive.
- If $$ k = 1 - \sqrt{7} $$, all terms become negative, which is still valid in a geometric sequence since the ratio remains consistent.

**Final Answer for Part (a):**
$$
k = 1 + \sqrt{7} \quad \text{or} \quad k = 1 - \sqrt{7}
$$

---

If you need assistance with parts (b) through (e) or with Problem 2, feel free to ask!
**(a) The value of $$ k $$ is $$ 1 + \sqrt{7} $$ or $$ 1 - \sqrt{7} $$.** **(b) The sequence is $$ (1 + \sqrt{7}) + 1, (1 + \sqrt{7}) - 1, 2(1 + \sqrt{7}) - 5, \ldots $$ or $$ (1 - \sqrt{7}) + 1, (1 - \sqrt{7}) - 1, 2(1 - \sqrt{7}) - 5, \ldots $$.** **(c) The general term of the sequence is $$ a_n = (1 + \sqrt{7}) \times r^{n-1} $$ or $$ a_n = (1 - \sqrt{7}) \times r^{n-1} $$, where $$ r $$ is the common ratio.** **(d) The 10th term is $$ a_{10} = (1 + \sqrt{7}) \times r^9 $$ or $$ a_{10} = (1 - \sqrt{7}) \times r^9 $$.** **(e) The term $$ -59049 $$ is the $$ n $$-th term where $$ n $$ is determined by solving $$ a_n = -59049 $$ using the general term formula.** --- **Problem 2:** **(a) Determine an expression for the $$ n^{\text{th}} $$ term of a geometric sequence where the 3rd term is -20 and the 6th term is 160.** **Given:** - $$ a_3 = -20 $$ - $$ a_6 = 160 $$ **Recall the formula for the $$ n^{\text{th}} $$ term of a geometric sequence:** $$ a_n = a_1 \times r^{n-1} $$ **Express $$ a_3 $$ and $$ a_6 $$ using the formula:** $$ a_3 = a_1 \times r^2 = -20 \quad \text{(1)} $$ $$ a_6 = a_1 \times r^5 = 160 \quad \text{(2)} $$ **Divide equation (2) by equation (1) to eliminate $$ a_1 $$:** $$ \frac{a_6}{a_3} = \frac{a_1 \times r^5}{a_1 \times r^2} = r^3 = \frac{160}{-20} = -8 $$ $$ r^3 = -8 $$ $$ r = \sqrt[3]{-8} = -2 $$ **Now, substitute $$ r = -2 $$ back into equation (1) to find $$ a_1 $$:** $$ a_1 \times (-2)^2 = -20 $$ $$ a_1 \times 4 = -20 $$ $$ a_1 = \frac{-20}{4} = -5 $$ **Therefore, the $$ n^{\text{th}} $$ term is:** $$ a_n = -5 \times (-2)^{n-1} $$ **Alternatively, using the general formula:** $$ a_n = a_1 \times r^{n-1} = -5 \times (-2)^{n-1} $$ **Final Answer:** $$ a_n = -5 \times (-2)^{n-1} $$ --- If you need further assistance with any part of these solutions, feel free to ask!

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