The following geometric sequence is given: $$ 10 ; 5 ; 2,5 ; 1.25 ; \ldots $$ 1.1 Calculate the value of the $$ 5^{ ext {th }} $$ term, $$ T_{5} $$, of this sequence. 1.2 Determine the $$ n $$ th, $$ T_{n} $$, in terms of $$ n $$ 1.3 Explain why the infinite series $$ 10+5+2,5+1,25+\cdots $$ conver 1.4 Determine $$ S_{\infty}-S_{n} $$ in the form $$ a b^{n} $$, where $$ S_{n} $$ is the sum of the sequence

Question

The following geometric sequence is given: $$ 10 ; 5 ; 2,5 ; 1.25 ; \ldots $$ 1.1 Calculate the value of the $$ 5^{	ext {th }} $$ term, $$ T_{5} $$, of this sequence. 1.2 Determine the $$ n $$ th, $$ T_{n} $$, in terms of $$ n $$ 1.3 Explain why the infinite series $$ 10+5+2,5+1,25+\cdots $$ conver 1.4 Determine $$ S_{\infty}-S_{n} $$ in the form $$ a b^{n} $$, where $$ S_{n} $$ is the sum of the sequence

UpStudy AI · Accepted Answer

To solve the problem step by step, let's first identify the known conditions of the geometric sequence given:

The first term $$ T_1 = 10 $$ and the second term $$ T_2 = 5 $$. The common ratio $$ r $$ can be calculated as follows:

$$
r = \frac{T_2}{T_1} = \frac{5}{10} = 0.5
$$

Now, let's proceed with each part of the problem.

### 1.1 Calculate the value of the $$ 5^{\text{th}} $$ term, $$ T_{5} $$

The formula for the $$ n $$-th term of a geometric sequence is given by:

$$
T_n = T_1 \cdot r^{n-1}
$$

Substituting the known values for $$ T_1 $$, $$ r $$, and $$ n = 5 $$:

$$
T_5 = 10 \cdot (0.5)^{5-1} = 10 \cdot (0.5)^4
$$

Calculating $$ (0.5)^4 $$:

$$
(0.5)^4 = \frac{1}{16}
$$

Now substituting back:

$$
T_5 = 10 \cdot \frac{1}{16} = \frac{10}{16} = \frac{5}{8}
$$

### 1.2 Determine the $$ n $$-th term, $$ T_{n} $$, in terms of $$ n $$

Using the formula for the $$ n $$-th term:

$$
T_n = 10 \cdot (0.5)^{n-1}
$$

### 1.3 Explain why the infinite series $$ 10 + 5 + 2.5 + 1.25 + \cdots $$ converges

An infinite geometric series converges if the absolute value of the common ratio $$ r $$ is less than 1. In this case, we have:

$$
|r| = 0.5 < 1
$$

Since the common ratio is less than 1, the series converges.

### 1.4 Determine $$ S_{\infty} - S_{n} $$ in the form $$ a b^{n} $$, where $$ S_{n} $$ is the sum of the sequence

The sum of the first $$ n $$ terms of a geometric series is given by:

$$
S_n = \frac{T_1(1 - r^n)}{1 - r}
$$

Substituting the known values:

$$
S_n = \frac{10(1 - (0.5)^n)}{1 - 0.5} = \frac{10(1 - (0.5)^n)}{0.5} = 20(1 - (0.5)^n)
$$

The sum of the infinite series $$ S_{\infty} $$ is given by:

$$
S_{\infty} = \frac{T_1}{1 - r} = \frac{10}{1 - 0.5} = 20
$$

Now, we can find $$ S_{\infty} - S_n $$:

$$
S_{\infty} - S_n = 20 - 20(1 - (0.5)^n) = 20 - 20 + 20(0.5)^n = 20(0.5)^n
$$

Thus, we have:

$$
S_{\infty} - S_n = 20(0.5)^n
$$

### Summary of Results
1. $$ T_5 = \frac{5}{8} $$
2. $$ T_n = 10 \cdot (0.5)^{n-1} $$
3. The series converges because $$ |r| < 1 $$.
4. $$ S_{\infty} - S_n = 20(0.5)^n $$
1. The 5th term $$ T_5 $$ is $$ \frac{5}{8} $$. 2. The $$ n $$-th term $$ T_n $$ is $$ 10 \times (0.5)^{n-1} $$. 3. The series converges because the common ratio is less than 1. 4. $$ S_{\infty} - S_n = 20 \times (0.5)^n $$.

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