Infinite Geometric Series

Question 1 of 8

A kangaroo slows down its jumping as it approaches a lake.
Starting at

$6$ m, its next jump covers half of the previous jump. Find the limiting sum of the distances that the kangaroo jumps over.

$S_∞=$ (12) $\text(m)$

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

Common Ratio Formula

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

First, solve for the value of $r$ .

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

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

	$=$	$1/2$	Simplify

Next, substitute the known values to the formula

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

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

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{6}}{1-\color{#00880A}{\frac{1}{2}}}$	Substitute known values

	$=$	$6/(1/2)$	Evaluate

	$=$	$12$

$S_∞=12 \text(m)$

Question 2 of 8

Find the limiting sum

$1-1/3+1/9-1/27…$

Write fractions as “a/b”

$S_∞=$ (3/4)

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

Common Ratio Formula

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

First, solve for the value of $r$ .

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

	$=$	$\frac{-\frac{1}{3}}{1}$	Substitute the first and second term

	$=$	$-1/3$

Next, substitute the known values to the formula

$\text(First term)$ $[a]$	$=$	$1$

$\text(Common Ratio)$ $[r]$	$=$	$-1/3$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{1}}{1-\left(\color{#00880A}{-\frac{1}{3}}\right)}$	Substitute known values

	$=$	$1/(4/3)$	Evaluate

	$=$	$3/4$

$S_∞=3/4$

Question 3 of 8

Find the value of

$r$ given that:

$S_∞=18$

$U_1=6$

Write fractions as “a/b”

$r=$ (2/3)

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

Substitute the known values then solve for $r$

$S_∞$	$=$	$18$
$\text(1st term)$	$=$	$6$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{18}$	$=$	$\frac{\color{#e65021}{6}}{1-\color{#00880A}{r}}$	Substitute known values

$18(1-r)$	$=$	$6$	Cross multiply
$18-18r$	$=$	$6$	Distribute
$18-18r$ $-18$	$=$	$6$ $-18$	Subtract $18$ from both sides
$-18r$ $divide(-18)$	$=$	$-12$ $divide(-18)$	Divide both sides by $-18$

$r$	$=$	$12/18$

$r$	$=$	$2/3$	Simplify

$r=2/3$

Question 4 of 8

Given that

$S_∞=4/5$ , find the value of

$x$

$x+x^2+x^3…$

Write fractions as “a/b”

$x=$ (4/9)

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

Common Ratio Formula

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

First, solve for the value of $r$ .

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

	$=$	$\frac{x^2}{x}$	Substitute the first and second term

	$=$	$x$

Next, substitute the known values then solve for

$x$

$S_∞$	$=$	$4/5$

$\text(First term)$ $[a]$	$=$	$x$
$\text(Common Ratio)$ $[r]$	$=$	$x$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{\frac{4}{5}}$	$=$	$\frac{\color{#e65021}{x}}{1-\color{#00880A}{x}}$	Substitute known values

$4(1-x)$	$=$	$5x$	Cross multiply
$4-4x$	$=$	$5x$	Distribute
$4-4x$ $+4x$	$=$	$5x$ $+4x$	Add $4x$ to both sides
$4$ $divide9$	$=$	$9x$ $divide9$	Divide both sides by $9$

$4/9$	$=$	$x$

$x$	$=$	$4/9$

$x=4/9$

Question 5 of 8

An acrobatic stunt plane flies vertically upwards. The further it goes up, the more it allows gravity to take over its flight, slowing it down until it reaches a peak point. At the first second, it travels

$120$ m,

$40$ m at the next, then

$13 1/3$ m, and so on. Find the total distance that the plane travels upwards until reaching the peak point.

$S_∞=$ (180) $\text(m)$

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

Common Ratio Formula

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

First, solve for the value of $r$ .

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

	$=$	$\frac{40}{120}$	Substitute the first and second term

	$=$	$1/3$	Simplify

Next, substitute the known values to the formula

$\text(First term)$ $[a]$	$=$	$120$

$\text(Common Ratio)$ $[r]$	$=$	$1/3$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{120}}{1-\color{#00880A}{\frac{1}{3}}}$	Substitute known values

	$=$	$120/(2/3)$	Evaluate

	$=$	$180$

$S_∞=180 \text(m)$

Question 6 of 8

A ball was dropped at a height of

$10 \text(m)$ . Find the limiting sum if the ball bounces back up by

$\text(4/5)$ of that height.

$S_∞=$ (90) $\text(m)$

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

First, find the value of the first term (

$a$ ).

The first term would be the first time the ball bounces, which is

$\text(4/5)$ of

$10$ .

$10xx4/5$

$=$

$8$

Substitute the known values to the formula

Note that the ball will be bouncing up and down each time, so the value must be multiplied by

$2$ .

$r$	$=$	$4/5$

$a$	$=$	$8$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

	$=$	$2\left(\frac{\color{#e65021}{8}}{1-\color{#00880A}{\frac{4}{5}}}\right)$	Substitute known values

	$=$	$2\left(\frac{8}{\frac{1}{5}}\right)$	Evaluate

	$=$	$2(40)$
	$=$	$80$

Finally, add the height from where the ball was first dropped, which is

$10$ .

$S_∞$	$=$	$80$ $+10$
	$=$	$90$

$S_∞=90 \text(m)$

Question 7 of 8

A speeding car brakes and travels

$18$ m in the first second,

$4.5$ m in the second, and

$1.125$ m in the third. How far will the car travel until it completely stops?

$S_∞=$ (24) $\text(m)$

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

Common Ratio Formula

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

First, solve for the value of $r$ .

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

	$=$	$\frac{4.5}{18}$	Substitute the first and second term

	$=$	$1/4$	Simplify

Next, substitute the known values to the formula

$\text(First term)$ $[a]$	$=$	$18$

$\text(Common Ratio)$ $[r]$	$=$	$1/4$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{18}}{1-\color{#00880A}{\frac{1}{4}}}$	Substitute known values

	$=$	$18/(3/4)$	Evaluate

	$=$	$24$

$S_∞=24 \text(m)$

Question 8 of 8

A man is standing on the roof of a building and drops a ball which travels

$36$ m onto the ground below. As it touches the ground, it goes back up

$2/3$ of its previous height and continues to dribble. Find out how high the ball rises after its sixth bounce.

$(i)$ How high would the ball rise after its sixth bounce?

$(ii)$ How many bounces will it take for the ball to have a height of

$7 1/9$ m?

$(iii)$ Find the total sum of the heights of the ball until it stops

Round off answer to

$2$ decimal places when applicable

$(i)$ (3.16)m

$(ii)$ (4) bounces

$(iii)$ (180)m

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Common Ratio Formula

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

Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

General Rule of a Geometric Sequence

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

$(i)$ Find how high the ball would rise after its sixth bounce.

First, sketch the first few data on how the ball bounces.

Note that each succeeding height is computed by getting

$2/3$ of the previous height

Next, substitute the known values to the general rule

$\text(Number of terms)$ $[n]$	$=$	$6$
$\text(First term)$ $[a]$	$=$	$24$

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

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

$U_{\color{#9a00c7}{6}}$	$=$	$\color{#e65021}{24}\cdot\color{#00880A}{\frac{2}{3}}^{\color{#9a00c7}{6}-1}$	Substitute known values

	$=$	$24\cdot\frac{2}{3}^5$	Evaluate

	$=$	$24\cdot\frac{32}{243}$

	$=$	$768/243$

	$=$	$3.16$ m

The height of the ball after the sixth bounce is

$3.16$ m

$(ii)$ Find how many bounces it will take for the ball to have a height of

$7 1/9$ m.

Substitute the known values to the general rule

$\text(nth term)$ $[U_\color{#9a00c7}{n}]$	$=$	$7 1/9$

$\text(First term)$ $[a]$	$=$	$24$

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

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

$7 \frac{1}{9}$	$=$	$\color{#e65021}{24}\cdot\color{#00880A}{\frac{2}{3}}^{\color{#9a00c7}{n}-1}$	Substitute known values

$64/9$ *$1/24$**	$=$	$24\cdot\frac{2}{3}^{n-1}\color{#CC0000}{\cdot\frac{1}{24}}$	Multiply both sides by $1/24$

$8/27$	$=$	$\frac{2}{3}^{n-1}$

$\log\frac{8}{27}$	$=$	$\log{\frac{2}{3}^{n-1}}$	Take the logarithm of both sides

$\log\frac{8}{27}$	$=$	$(n-1)\log\frac{2}{3}$

$\frac{\log \frac{8}{27}}{\log \frac{2}{3}}$	$=$	$n-1$

$1+\frac{\log \frac{8}{27}}{\log \frac{2}{3}}$	$=$	$n$

$1+3$	$=$	$n$
$n$	$=$	$4$

It takes

$4$ bounces for the ball to have a height of

$7 1/9$ m

$(iii)$ Find the total sum of the heights of the ball until it stops.

First, substitute the known values to the limiting sum formula

$\text(First term)$ $[a]$	$=$	$24$

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

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{24}}{1-\color{#00880A}{\frac{2}{3}}}$	Substitute known values

	$=$	$24/(1/3)$	Evaluate

	$=$	$72$

This is just the total of the distances traveled by the ball from the air to the ground.

It does not include the distance when it bounces from the ground back to the air and the initial height of

$36$ m

Finally, double the limiting sum and add

$36$

	$2(72)+36$
$=$	$144+36$
$=$	$180$ m

The total sum of the heights of the ball until it stops is

$180$ m

$(i) 3.16$ m

$(ii) 4$ bounces

$(iii) 180$ m

Infinite Geometric Series

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Quiz summary

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1. Question

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Great Work!

Limiting Sum Formula

Common Ratio Formula

2. Question

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Excellent!

Limiting Sum Formula

Common Ratio Formula

3. Question

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Nice Job!

Limiting Sum Formula

4. Question

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Well Done!

Limiting Sum Formula

Common Ratio Formula

5. Question

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Correct!

Limiting Sum Formula

Common Ratio Formula

6. Question

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Limiting Sum Formula

7. Question

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Fantastic!

Limiting Sum Formula

Common Ratio Formula

8. Question

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Excellent!

Common Ratio Formula

Limiting Sum Formula

General Rule of a Geometric Sequence

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