Geometric Sequences

Question 1 of 3

Given the sequence

3, 6, 12, 24 \dots

$3,6,12,24…$ , find:

(i)

$(i)$ The

8 t h

$8th$ term

(i i)

$(ii)$ If

1536

$1536$ is part of the sequence

For part

i i

$ii$ , write

Y

$Y$ for yes and

N

$N$ for no

$(i)$ $(i)$ $U_{8} =$ $U_8=$ (384)

$(i i)$ $(ii)$ (Y, y)

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General Rule of a Geometric Sequence

U_{n} = a r^{n - 1}

$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$

Common Ratio Formula

r = \frac{U_{2}}{U_{1}} = \frac{U_{3}}{U_{2}}

$\color{#00880A}{r}=\frac{U_2}{U_1}=\frac{U_3}{U_2}$

(i)

$(i)$ Finding the

8 t h

$8th$ term

First, solve for the value of $r$ $r$ .

$\color{#00880A}{r}$	$=$ $=$	$\frac{U_2}{U_1}$

	$=$ $=$	$\frac{6}{3}$	Substitute the first and second term

	$=$ $=$	$2$ $2$

Next, substitute the known values to the general rule

$Number of terms$ $\text(Number of terms)$ $[n]$ $[n]$	$=$ $=$	$8$ $8$
$First term$ $\text(First term)$ $[a]$ $[a]$	$=$ $=$	$3$ $3$
$Common Ratio$ $\text(Common Ratio)$ $[r]$ $[r]$	$=$ $=$	$2$ $2$

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$

$U_{\color{#9a00c7}{8}}$	$=$ $=$	$\color{#e65021}{3}\cdot\color{#00880A}{2}^{\color{#9a00c7}{8}-1}$	Substitute known values

	$=$ $=$	$3\cdot(2^7)$	Evaluate
	$=$ $=$	$3\cdot128$
	$=$ $=$	$384$ $384$

(i i)

$(ii)$ Finding if

1536

$1536$ is part of the sequence

Next, substitute the known values to the general rule and solve for $n$ $n$

$Nth term$ $\text(Nth term)$ $[U_{n}]$ $[U_n]$	$=$ $=$	$1536$ $1536$
$First term$ $\text(First term)$ $[a]$ $[a]$	$=$ $=$	$3$ $3$
$Common Ratio$ $\text(Common Ratio)$ $[r]$ $[r]$	$=$ $=$	$2$ $2$

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$

$1536$	$=$ $=$	$\color{#e65021}{3}\cdot\color{#00880A}{2}^{\color{#9a00c7}{n}-1}$	Substitute known values

$1536$ $1536$ $\div 3$ $divide3$	$=$ $=$	$3 (2^{n - 1})$ $3(2^(n-1))$ $\div 3$ $divide3$	Divide both sides by $3$ $3$
$512$ $512$	$=$ $=$	$2^{n - 1}$ $2^(n-1)$
$2^{9}$ $2^9$	$=$ $=$	$2^{n - 1}$ $2^(n-1)$	$512 = 2^{9}$ $512=2^9$
$9$ $9$	$=$ $=$	$n - 1$ $n-1$	Equate the exponents
$9$ $9$ $+ 1$ $+1$	$=$ $=$	$n - 1$ $n-1$ $+ 1$ $+1$	Add $1$ $1$ to both sides
$10$ $10$	$=$ $=$	$n$ $n$
$n$ $n$	$=$ $=$	$10$ $10$

1536

$1536$ is the value of the

10 t h

$10th$ term. Therefore, it is a part of the sequence

(i) U_{8} = 384

$(i) U_8=384$

(i i) Yes

$(ii) \text(Yes)$

Question 2 of 3

Find the value of

n

$n$

6, 3, 1 \frac{1}{2} \dots \frac{3}{128}

$6,3,1 1/2…3/(128)$

$n =$ $n=$ (9)

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General Rule of a Geometric Sequence

U_{n} = a r^{n - 1}

$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$

Common Ratio Formula

r = \frac{U_{2}}{U_{1}} = \frac{U_{3}}{U_{2}}

$\color{#00880A}{r}=\frac{U_2}{U_1}=\frac{U_3}{U_2}$

First, solve for the value of $r$ $r$ .

$\color{#00880A}{r}$	$=$ $=$	$\frac{U_2}{U_1}$

	$=$ $=$	$\frac{3}{6}$	Substitute the first and second term

	$=$ $=$	$\frac{1}{2}$ $1/2$

Next, substitute the known values to the general rule and solve for $n$ $n$

$Nth term$ $\text(Nth term)$ $[U_{n}]$ $[U_n]$	$=$ $=$	$\frac{3}{128}$ $3/128$

$First term$ $\text(First term)$ $[a]$ $[a]$	$=$ $=$	$6$ $6$

$Common Ratio$ $\text(Common Ratio)$ $[r]$ $[r]$	$=$ $=$	$\frac{1}{2}$ $1/2$

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$

$\frac{3}{128}$	$=$ $=$	$\color{#e65021}{6}\cdot\color{#00880A}{\frac{1}{2}}^{\color{#9a00c7}{n}-1}$	Substitute known values

$\frac{3}{128}$ $3/128$ $\times \frac{1}{6}$ $times1/6$	$=$ $=$	$[6 {(\frac{1}{2})}^{n - 1}]$ $[6(1/2)^(n-1)]$ $\times \frac{1}{6}$ $times1/6$	Multiply both sides by $\frac{1}{6}$ $1/6$

$\frac{1}{256}$ $1/256$	$=$ $=$	${(\frac{1}{2})}^{n - 1}$ $(1/2)^(n-1)$	$\frac{6}{6} = 1$ $6/6=1$

$\frac{1}{2^{8}}$ $1/(2^8)$	$=$ $=$	$\frac{1}{2^{n - 1}}$ $1/(2^(n-1))$	$256 = 2^{8}$ $256=2^8$

$8$ $8$	$=$ $=$	$n - 1$ $n-1$	Equate the exponents
$8$ $8$ $+ 1$ $+1$	$=$ $=$	$n - 1$ $n-1$ $+ 1$ $+1$	Add $1$ $1$ to both sides
$9$ $9$	$=$ $=$	$n$ $n$
$n$ $n$	$=$ $=$	$9$ $9$

n = 9

$n=9$

Question 3 of 3

Find the

8 t h

$8th$ term

\frac{2}{7} + \frac{3}{7} + \frac{9}{14} \dots

$2/7+3/7+9/14…$

1. $5 \frac{53}{448}$ $5 53/448$
2. $4 \frac{395}{448}$ $4 395/448$
3. $4 \frac{448}{53}$ $4 448/53$
4. $4 \frac{448}{395}$ $4 448/395$

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General Rule of a Geometric Sequence

U_{n} = a r^{n - 1}

$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$

Common Ratio Formula

r = \frac{U_{2}}{U_{1}} = \frac{U_{3}}{U_{2}}

$\color{#00880A}{r}=\frac{U_2}{U_1}=\frac{U_3}{U_2}$

First, solve for the value of $r$ $r$ .

$\color{#00880A}{r}$	$=$ $=$	$\frac{U_2}{U_1}$

	$=$ $=$	$\frac{\frac{3}{7}}{\frac{2}{7}}$	Substitute the first and second term

	$=$ $=$	$\frac{3}{2}$ $3/2$

Next, substitute the known values to the general rule

$Number of terms$ $\text(Number of terms)$ $[n]$ $[n]$	$=$ $=$	$8$ $8$

$First term$ $\text(First term)$ $[a]$ $[a]$	$=$ $=$	$\frac{2}{7}$ $2/7$

$Common Ratio$ $\text(Common Ratio)$ $[r]$ $[r]$	$=$ $=$	$\frac{3}{2}$ $3/2$

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$

$U_{\color{#9a00c7}{8}}$	$=$ $=$	$\color{#e65021}{\frac{2}{7}}\cdot\color{#00880A}{\frac{3}{2}}^{\color{#9a00c7}{8}-1}$	Substitute known values

	$=$ $=$	$\frac{2}{7}\cdot\left(\frac{3}{2}\right)^2$	Evaluate

	$=$ $=$	$\frac{2}{7}\cdot\frac{2187}{128}$

	$=$ $=$	$\frac{2187}{448}$ $2187/448$

	$=$ $=$	$4 \frac{395}{448}$ $4 395/448$	Simplify

U_{8} = 4 \frac{395}{448}

$U_8=4 395/448$

Geometric Sequences

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Quiz summary

Information

1. Question

Hint

Correct!

General Rule of a Geometric Sequence

Common Ratio Formula

2. Question

Hint

Fantastic!

General Rule of a Geometric Sequence

Common Ratio Formula

3. Question

Hint

Excellent!

General Rule of a Geometric Sequence

Common Ratio Formula

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