$$ x-4 ; x+2 ; 3 x+1 ; \ldots $$ are the first three terms of a geometric sequence. Determine the sequence if $$ x $$ is positive. $$ t+1 ; 1-t ; 1-5 t ; \ldots $$ are the first three terms of a geometric sequence. (a) Determine the numerical value of $$ t $$ where $$ t
eq 0 $$. b) Determine the sequence. d) Determine the 10 th term. d) Which term equals $$ 10 \frac{2}{3} $$ ?

Question

$$ x-4 ; x+2 ; 3 x+1 ; \ldots $$ are the first three terms of a geometric sequence. Determine the sequence if $$ x $$ is positive. $$ t+1 ; 1-t ; 1-5 t ; \ldots $$ are the first three terms of a geometric sequence. (a) Determine the numerical value of $$ t $$ where $$ t 
eq 0 $$. b) Determine the sequence. d) Determine the 10 th term. d) Which term equals $$ 10 \frac{2}{3} $$ ?

UpStudy AI · Accepted Answer

Let's solve each part of the problem step by step.

### Given:
The first three terms of a geometric sequence are:
$$ t + 1, \quad 1 - t, \quad 1 - 5t, \ldots $$

### (a) Determine the numerical value of $$ t $$ where $$ t \neq 0 $$.

In a geometric sequence, the ratio between consecutive terms is constant. Therefore:

$$
\frac{1 - t}{t + 1} = \frac{1 - 5t}{1 - t}
$$

Cross-multiplying gives:

$$
(1 - t)^2 = (t + 1)(1 - 5t)
$$

Expanding both sides:

$$
1 - 2t + t^2 = t \cdot 1 + t \cdot (-5t) + 1 \cdot 1 + 1 \cdot (-5t)
$$
$$
1 - 2t + t^2 = t - 5t^2 + 1 - 5t
$$
$$
1 - 2t + t^2 = 1 - 4t - 5t^2
$$

Bringing all terms to one side:

$$
t^2 + 5t^2 - 2t + 4t = 0
$$
$$
6t^2 + 2t = 0
$$
$$
2t(3t + 1) = 0
$$

Solutions:
$$
t = 0 \quad \text{or} \quad t = -\frac{1}{3}
$$

Since $$ t \neq 0 $$, we have:
$$
t = -\frac{1}{3}
$$

### (b) Determine the sequence.

With $$ t = -\frac{1}{3} $$, let's compute the first few terms:

1. **First term ($$ a_1 $$)**:
   $$
   t + 1 = -\frac{1}{3} + 1 = \frac{2}{3}
   $$
2. **Second term ($$ a_2 $$)**:
   $$
   1 - t = 1 - \left(-\frac{1}{3}\right) = 1 + \frac{1}{3} = \frac{4}{3}
   $$
3. **Third term ($$ a_3 $$)**:
   $$
   1 - 5t = 1 - 5\left(-\frac{1}{3}\right) = 1 + \frac{5}{3} = \frac{8}{3}
   $$

**Common Ratio ($$ r $$)**:
$$
r = \frac{a_2}{a_1} = \frac{\frac{4}{3}}{\frac{2}{3}} = 2
$$

So, the sequence is:
$$
\frac{2}{3}, \quad \frac{4}{3}, \quad \frac{8}{3}, \quad \frac{16}{3}, \quad \frac{32}{3}, \quad \ldots
$$

Or in general terms:
$$
a_n = \frac{2}{3} \times 2^{n-1}
$$

### (c) Determine the 10th term.

Using the general formula:
$$
a_{10} = \frac{2}{3} \times 2^{10-1} = \frac{2}{3} \times 2^9 = \frac{2}{3} \times 512 = \frac{1024}{3} = 341 \frac{1}{3}
$$

### (d) Which term equals $$ 10 \frac{2}{3} $$?

First, convert $$ 10 \frac{2}{3} $$ to an improper fraction:
$$
10 \frac{2}{3} = \frac{32}{3}
$$

Set the general term equal to $$ \frac{32}{3} $$:
$$
\frac{2}{3} \times 2^{n-1} = \frac{32}{3}
$$

Divide both sides by $$ \frac{2}{3} $$:
$$
2^{n-1} = 16
$$

Since $$ 16 = 2^4 $$:
$$
n - 1 = 4 \quad \Rightarrow \quad n = 5
$$

**Therefore, the 5th term of the sequence is $$ 10 \frac{2}{3} $$.**

### Summary of Answers:

**(a)** $$ t = -\frac{1}{3} $$

**(b)** The sequence is $$ \frac{2}{3}, \frac{4}{3}, \frac{8}{3}, \frac{16}{3}, \frac{32}{3}, \ldots $$ with common ratio 2.

**(c)** The 10th term is $$ \frac{1024}{3} $$ or $$ 341 \frac{1}{3} $$.

**(d)** The 5th term is $$ \frac{32}{3} $$ or $$ 10 \frac{2}{3} $$.
$$ t = -\frac{1}{3} $$ The sequence is $$ \frac{2}{3}, \frac{4}{3}, \frac{8}{3}, \frac{16}{3}, \frac{32}{3}, \ldots $$ with a common ratio of 2. The 10th term is $$ \frac{1024}{3} $$ or $$ 341 \frac{1}{3} $$. The 5th term is $$ \frac{32}{3} $$ or $$ 10 \frac{2}{3} $$.

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