QUESTION 2 Given the geometric sequence: $$ \frac{9}{2} ; 9 ; 18 ; \ldots ; 2304 $$ 2.1.1 Determine the value of $$ r $$, the common ratio. 2.1.2 How many terms are there in the sequence? Given: $$ \sum_{k=1}^{\infty} 6(m)^{k-1}=12 $$. Determine the value of $$ m $$. The $$ 3^{ ext {rd }} $$ term of a geometric series is 18 and the $$ 5^{ ext {th }} $$ term is 162 . Determine the sum of the first 7 terms, where $$ r

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QUESTION 2 Given the geometric sequence: $$ \frac{9}{2} ; 9 ; 18 ; \ldots ; 2304 $$ 2.1.1 Determine the value of $$ r $$, the common ratio. 2.1.2 How many terms are there in the sequence? Given: $$ \sum_{k=1}^{\infty} 6(m)^{k-1}=12 $$. Determine the value of $$ m $$. The $$ 3^{	ext {rd }} $$ term of a geometric series is 18 and the $$ 5^{	ext {th }} $$ term is 162 . Determine the sum of the first 7 terms, where $$ r<0 $$. The general term of a quadratic number pattern is $$ T_{n}=a n^{2}+b n+c $$ and its first term is 8 . The general term of the first differences of the pattern is $$ t_{k}=4 k-2 $$. 2.4.1 Determine the next TWO terms of the quadratic number pattern $$ T_{n} $$. 2.4) 2.4.2 Hence, or otherwise, show that the general term of the quadratic number pattern is given by $$ T_{n}=2 n^{2}-4 n+10 $$. 2.4.3 Which term of the quadratic number pattern will be equal to 3050 ? 2. 2. 2.

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### 2.1.1 Determine the value of $$ r $$, the common ratio.

In a geometric sequence, the common ratio $$ r $$ can be found by dividing any term by the previous term.

Given the first two terms:
- First term $$ a_1 = \frac{9}{2} $$
- Second term $$ a_2 = 9 $$

The common ratio $$ r $$ is calculated as follows:

$$
r = \frac{a_2}{a_1} = \frac{9}{\frac{9}{2}} = 9 \times \frac{2}{9} = 2
$$

### 2.1.2 How many terms are there in the sequence?

The $$ n $$-th term of a geometric sequence can be expressed as:

$$
a_n = a_1 \cdot r^{n-1}
$$

We know the last term $$ a_n = 2304 $$, $$ a_1 = \frac{9}{2} $$, and $$ r = 2 $$. We can set up the equation:

$$
2304 = \frac{9}{2} \cdot 2^{n-1}
$$

To find $$ n $$, we first multiply both sides by 2:

$$
4608 = 9 \cdot 2^{n-1}
$$

Now, divide both sides by 9:

$$
\frac{4608}{9} = 2^{n-1}
$$

Calculating $$ \frac{4608}{9} $$:

$$
\frac{4608}{9} = 512
$$

Now we have:

$$
2^{n-1} = 512
$$

Since $$ 512 = 2^9 $$, we can equate the exponents:

$$
n - 1 = 9 \implies n = 10
$$

### Given: $$ \sum_{k=1}^{\infty} 6(m)^{k-1}=12 $$. Determine the value of $$ m $$.

The sum of an infinite geometric series is given by:

$$
S = \frac{a}{1 - r}
$$

where $$ a $$ is the first term and $$ r $$ is the common ratio. Here, $$ a = 6 $$ and $$ r = m $$:

$$
12 = \frac{6}{1 - m}
$$

Cross-multiplying gives:

$$
12(1 - m) = 6
$$

Expanding and rearranging:

$$
12 - 12m = 6 \implies 12m = 6 \implies m = \frac{1}{2}
$$

### The $$ 3^{\text{rd}} $$ term of a geometric series is 18 and the $$ 5^{\text{th}} $$ term is 162. Determine the sum of the first 7 terms, where $$ r < 0 $$.

Let the first term be $$ a $$ and the common ratio be $$ r $$. The terms can be expressed as:

$$
a \cdot r^2 = 18 \quad (1)
$$
$$
a \cdot r^4 = 162 \quad (2)
$$

Dividing equation (2) by equation (1):

$$
\frac{a \cdot r^4}{a \cdot r^2} = \frac{162}{18} \implies r^2 = 9 \implies r = 3 \text{ or } r = -3
$$

Since $$ r < 0 $$, we have $$ r = -3 $$.

Substituting $$ r $$ back into equation (1):

$$
a \cdot (-3)^2 = 18 \implies a \cdot 9 = 18 \implies a = 2
$$

Now, we can find the sum of the first 7 terms:

$$
S_7 = a \frac{1 - r^7}{1 - r} = 2 \frac{1 - (-3)^7}{1 - (-3)}
$$

Calculating $$ (-3)^7 = -2187 $$:

$$
S_7 = 2 \frac{1 + 2187}{1 + 3} = 2 \frac{2188}{4} = 2 \cdot 547 = 1094
$$

### 2.4.1 Determine the next TWO terms of the quadratic number pattern $$ T_{n} $$.

Given $$ T_{n} = a n^2 + b n + c $$ and the first term $$ T_1 = 8 $$:

$$
T_1 = a(1)^2 + b(1) + c = 8 \quad (3)
$$

The first differences are given by $$ t_k = 4k - 2 $$. The second differences are constant, indicating a quadratic pattern.

The first differences for the first few terms are:

- $$ T_2 - T_1 = t_1 = 2 $$
- $$ T_3 - T_2 = t_2 = 6 $$
- $$ T_4 - T_3 = t_3 = 10 $$

The second differences are constant at 4, confirming $$ a = 2 $$.

Using $$ a = 2 $$ in equation (3):

$$
2(1)^2 + b(1) + c = 8 \implies 2 + b + c = 8 \implies b + c = 6 \quad (4)
$$

Now, we can find $$ T_2 $$ and $$ T_3 $$:

$$
T_2 = T_1 + t_1 = 8 + 2 = 10
$$
$$
T_3 = T_2 + t_2 = 10 + 6 = 16
$$

Thus, the next two terms are $$ 10 $$ and $$ 16 $$.

### 2.4.2 Hence, or otherwise, show that the general term of the quadratic number pattern is given by $$ T_{n}=2 n^{2}-4 n+10 $$.

We have $$ a = 2 $$. Now we need to find $$ b $$ and $$ c $$. From equation (4):

Assuming $$ c = 10 $$ (as $$ T_1 = 8 $$ and $$ T_2 = 10 $$), we can find $$ b $$:

$$
b + 10 = 6 \implies b = -4
$$

Thus, the general term is:

$$
T_n = 2n^2 - 4n + 10
$$

### 2.4.3 Which term of the quadratic number pattern will be equal to 3050?

Setting $$ T_n = 3050 $$:

$$
2n^2 - 4n + 10 = 3050
$$

Rearranging gives:

\[
2n
**Solutions:** 1. **Common Ratio ($$ r $$)**: - $$ r = 2 $$ 2. **Number of Terms**: - There are 10 terms in the sequence. 3. **Value of $$ m $$**: - $$ m = \frac{1}{2} $$ 4. **Sum of the First 7 Terms**: - The sum is 1094. 5. **Next Two Terms of the Quadratic Pattern**: - $$ T_2 = 10 $$ - $$ T_3 = 16 $$ 6. **General Term of the Quadratic Pattern**: - $$ T_n = 2n^2 - 4n + 10 $$ 7. **Term Equal to 3050**: - The term is $$ T_{20} = 3050 $$.

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