3.2 The first three terms of an arithmetic geometric sequence are 16,8 and 4 respectively. 3.2.1 Determine the $$ n^{ ext {th }} $$ term of the sequence 3.2.2 Determine all possible values of $$ n $$ for which the sum of the first $$ n $$ terms of this sequence is greater than 31 3.2.3 Calculate the sum to infinity of this sequence

Question

3.2 The first three terms of an arithmetic geometric sequence are 16,8 and 4 respectively. 3.2.1 Determine the $$ n^{	ext {th }} $$ term of the sequence 3.2.2 Determine all possible values of $$ n $$ for which the sum of the first $$ n $$ terms of this sequence is greater than 31 3.2.3 Calculate the sum to infinity of this sequence

UpStudy AI · Accepted Answer

To solve the problem, we need to analyze the arithmetic geometric sequence given by the first three terms: 16, 8, and 4.

### Step 1: Identify the sequence characteristics

1. **First term** ($$ a_1 $$): 16
2. **Second term** ($$ a_2 $$): 8
3. **Third term** ($$ a_3 $$): 4

### Step 2: Determine the common ratio and common difference

In an arithmetic geometric sequence, each term is formed by multiplying the previous term by a common ratio and then adding a common difference.

1. **Common ratio** ($$ r $$):
   $$
   r = \frac{a_2}{a_1} = \frac{8}{16} = \frac{1}{2}
   $$

2. **Common difference** ($$ d $$):
   $$
   d = a_2 - a_1 = 8 - 16 = -8
   $$

### Step 3: Formulate the $$ n^{\text{th}} $$ term

The $$ n^{\text{th}} $$ term of an arithmetic geometric sequence can be expressed as:
$$
a_n = a_1 \cdot r^{n-1} + d \cdot \frac{1 - r^{n-1}}{1 - r}
$$

Substituting the known values:
- $$ a_1 = 16 $$
- $$ r = \frac{1}{2} $$
- $$ d = -8 $$

We can now calculate $$ a_n $$.

### Step 4: Calculate $$ a_n $$

Let's calculate $$ a_n $$:
$$
a_n = 16 \cdot \left(\frac{1}{2}\right)^{n-1} - 8 \cdot \frac{1 - \left(\frac{1}{2}\right)^{n-1}}{1 - \frac{1}{2}}
$$
$$
= 16 \cdot \left(\frac{1}{2}\right)^{n-1} - 8 \cdot \frac{1 - \left(\frac{1}{2}\right)^{n-1}}{\frac{1}{2}}
$$
$$
= 16 \cdot \left(\frac{1}{2}\right)^{n-1} - 16 \cdot \left(1 - \left(\frac{1}{2}\right)^{n-1}\right)
$$
$$
= 16 \cdot \left(\frac{1}{2}\right)^{n-1} - 16 + 16 \cdot \left(\frac{1}{2}\right)^{n-1}
$$
$$
= 32 \cdot \left(\frac{1}{2}\right)^{n-1} - 16
$$

### Step 5: Solve for $$ n^{\text{th}} $$ term

Now, we can express the $$ n^{\text{th}} $$ term:
$$
a_n = 32 \cdot \left(\frac{1}{2}\right)^{n-1} - 16
$$

### Step 6: Determine the sum of the first $$ n $$ terms

The sum of the first $$ n $$ terms of an arithmetic geometric sequence can be calculated using the formula:
$$
S_n = \frac{a_1(1 - r^n)}{1 - r} + d \cdot \frac{1 - r^n}{(1 - r)^2}
$$

Substituting the known values:
$$
S_n = \frac{16(1 - \left(\frac{1}{2}\right)^n)}{1 - \frac{1}{2}} + (-8) \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\left(\frac{1}{2}\right)^2}
$$
$$
= 32(1 - \left(\frac{1}{2}\right)^n) - 32(1 - \left(\frac{1}{2}\right)^n)
$$
$$
= 32(1 - \left(\frac{1}{2}\right)^n) - 32(1 - \left(\frac{1}{2}\right)^n)
$$
$$
= 32(1 - \left(\frac{1}{2}\right)^n) - 32 + 32 \cdot \left(\frac{1}{2}\right)^n
$$
$$
= 32 - 32 \cdot \left(\frac{1}{2}\right)^n
$$

### Step 7: Solve for $$ n $$ such that $$ S_n > 31 $$

We need to find $$ n $$ such that:
$$
32 - 32 \cdot \left(\frac{1}{2}\right)^n > 31
$$
$$
-32 \cdot \left(\frac{1}{2}\right)^n > -1
$$
$$
\left(\frac{1}{2}\right)^n < \frac{1}{32}
$$
Taking logarithm base 2:
$$
n > 5
$$

### Step 8: Calculate the sum to infinity

The sum to infinity $$ S $$ of an arithmetic geometric sequence is given by:
$$
S = \frac{a_1}{1 - r} + \frac{d}{(1 - r)^2}
$$
Substituting the known values:
$$
S = \frac{16}{1 - \frac{1}{2}} + \frac{-8}{(1 - \frac{1}{2})^2}
$$
$$
= 32 - 32 = 0
$$

### Final Answers

1. **$$ n^{\text{th}} $$ term**: $$ a_n = 32 \cdot \left(\frac{1}{2}\right)^{n-1} - 16 $$
2. **Values of $$ n $$ for $$ S_n > 31 $$**: $$ n > 5 $$
3. **Sum to infinity**: $$ S = 0 $$
1. **$$ n^{\text{th}} $$ term**: $$ a_n = 32 \cdot \left(\frac{1}{2}\right)^{n-1} - 16 $$ 2. **Values of $$ n $$**: $$ n > 5 $$ 3. **Sum to infinity**: $$ S = 0 $$

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