Question

Determine the common ratio and the next 3 terms of the following geometric
sequence. Write your answer on the space provided. Sample way to type the answer; r= 2 (space) n= 1, 2, 3
Note: Incase the answer is fraction type in this way: 1/2
then upload your solution at the end of the written work.

1. 972, 324, 108, …

2. -3, 12, -48, …

3. 0.1, 0.5, 2.5, …

4. 10 000, 1 000, 100, …

5. -4, 4, -4, ...

6. Type the first 2 terms of the geometric sequence having a1= 2 r= 4 Type in this way: n= 1, 2 (with space)

7. Type the first 2 terms of the geometric sequence having a1= 5 r= -1/3 Type in this way: n= 1, 2 (with space)

8. Find the nth term of the geometric sequence whose initial term a1 and common ratio r are given. Then find the indicated term of the sequence. a1= 3 r= 5 find 7th term. Type your answer in this way: an= 200 (with space)

9. Find the nth term of the geometric sequence whose initial term a1 and common ratio r are given. Then find the indicated term of the sequence. a1= 0 r= 4 find 30th term. Type your answer in this way: an= 200 (with space)

10. In the sequence 1, 2, 4, 8, ... which term is 64? Type your answer in this way: 4th term (With space)

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Answer 1

A geometric sequence is characterized with a common ratio

1. $\mathbf{r = \frac{1}{3}\ and\ \mathbf{n = 36, 12, 4}}$ ,
2. $\mathbf{r = -4 \ and\ n = 192, -768, 3072}$,
3. $\mathbf{r = 5 \ and\ n = 12.5, 62.5, 312.5}$,
4. $\mathbf{r = \frac{1}{10} \ and\ n = 10, 1, \frac{1}{10}}$,
5. $\mathbf{r = -1 \ and\ n = 4,-4,4}$,
6. $\mathbf{n = 2, 8}$,
7. $\mathbf{n = 5, -\frac 53}$,
8. $\mathbf{a_7 = 46875}$,
9. $\mathbf{a_{30} = 0}$,
10. 64 is 7th term

The common ratio and the next three terms

Given the first three terms, the common ratio (r) is:

$\mathbf{r = \frac{T_2}{T_1}}$

While the nth term is:

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

So, we have:

1. 972, 324, 108, …

The common ratio (r) is:

$\mathbf{r = \frac{324}{972}}$

$\mathbf{r = \frac{1}{3}}$

The next three terms are:

$\mathbf{T_4 = 972 \times (\frac 13)^3 = 36}$

$\mathbf{T_5 = 972 \times (\frac 13)^4 = 12}$

$\mathbf{T_6 = 972 \times (\frac 13)^5 = 4}$

So,

$\mathbf{r = \frac{1}{3}\ and\ \mathbf{n = 36, 12, 4}}$

2. -3, 12, -48, …

The common ratio (r) is:

$\mathbf{r = \frac{12}{-3}}$

$\mathbf{r = -4}$

The next three terms are:

$\mathbf{T_4 = -3 \times (-4)^3 = 192}$

$\mathbf{T_5 = -3 \times (-4)^4 = -768}$

$\mathbf{T_6 = -3 \times (-4)^5 = 3072}$

So,

$\mathbf{r = -4 \ and\ n = 192, -768, 3072}$

3. 0.1, 0.5, 2.5, …

The common ratio (r) is:

$\mathbf{r = \frac{0.5}{0.1}}$

$\mathbf{r = 5}$

The next three terms are:

$\mathbf{T_4 = 0.1 \times (5)^3 = 12.5}$

$\mathbf{T_5 = 0.1 \times (5)^4 = 62.5}$

$\mathbf{T_6 = 0.1 \times (5)^5 = 312.5}$

So,

$\mathbf{r = 5 \ and\ n = 12.5, 62.5, 312.5}$

4. 10 000, 1 000, 100, …

The common ratio (r) is:

$\mathbf{r = \frac{1000}{10000}}$

$\mathbf{r = \frac{1}{10}}$

The next three terms are:

$\mathbf{T_4 = 10000 \times (\frac{1}{10})^3 = 10}$

$\mathbf{T_5 = 10000 \times (\frac{1}{10})^4 = 1}$

$\mathbf{T_6 = 10000 \times (\frac{1}{10})^5 = \frac{1}{10}}$

So,

$\mathbf{r = \frac{1}{10} \ and\ n = 10, 1, \frac{1}{10}}$

5. -4, 4, -4, ...

The common ratio (r) is:

$\mathbf{r = \frac{4}{-4}}$

$\mathbf{r = -1}$

The next three terms are:

$\mathbf{T_4 = -4 \times (-1)^3 = 4}$

$\mathbf{T_5 = -4 \times (-1)^4 = -4}$

$\mathbf{T_6 = -4 \times (-1)^5 = 4}$

So,

$\mathbf{r = -1 \ and\ n = 4,-4,4}$

The first 2 terms

$6.\ \mathbf{a_1 =2 , r = 4}$

The nth term is:

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

So, the second term is:

$\mathbf{a_2 = 2 \times 4 = 8}$

So, we have:

$\mathbf{n = 2, 8}$

$7.\ \mathbf{a_1 =5, r = -\frac 13}$

The nth term is:

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

So, the second term is:

$\mathbf{a_2 = 5 \times -\frac 13 = -\frac 53}$

So, we have:

$\mathbf{n = 5, -\frac 53}$

The nth term

$\mathbf{8.\ a_1 = 3, r = 5}$

The 7th is calculated using:

$\mathbf{a_7 = a_1 \times r^6}$

$\mathbf{a_{7} = 3 \times 5^{6}}$

$\mathbf{a_7 = 46875}$

$\mathbf{9.\ a_1 = 0, r = 4}$

The 30th is calculated using:

$\mathbf{a_{30} = a_1 \times r^{29}}$

$\mathbf{a_{30} = 0 \times 4^{29}}$

$\mathbf{a_{30} = 0}$

The number of term

$\mathbf{10.\ n = 1,2,4,8....}$

$\mathbf{a_n = 64}$

$\mathbf{a_1 = 1}$

Start by calculating the common ratio (r)

$\mathbf{r = \frac 21}$

$\mathbf{r = 2}$

So, the nth term is:

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

Substitute known values

$\mathbf{64 = 1 \times 2^{n-1}}$

$\mathbf{64 = 2^{n-1}}$

Express 64 as 2^6

$\mathbf{2^6 = 2^{n-1}}$

Cancel out 2 on both sides

$\mathbf{6 = n-1}$

Add 1 to both sides

$\mathbf{7 = n}$

Rewrite as:

$\mathbf{n = 7}$

Hence:

64 is 7th term

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