Back to the on-screen lesson ·
Sum finite and infinite geometric series, including a savings plan, and read percentages off a normal distribution with the empirical rule and z-scores.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you learn to add a geometric sequence without adding it term by term: one shift-and-subtract argument turns the sum of the first $n$ terms into a formula, and letting $n$ run away turns it into a formula for an infinite sum whenever the common ratio is smaller than $1$ in size — which is what makes a savings plan computable. Then you meet the normal distribution, the bell-shaped model behind heights, measurement errors and machine tolerances, and read percentages off it with the empirical rule and with $z$-scores.
You already recognise a geometric sequence: each term is the one before multiplied by a fixed number, so $3, 6, 12, 24, \ldots$ multiplies by $2$ every step and the $n$th term is $ar^{n-1}$. You already know that adding $5\%$ interest means multiplying by $1.05$, and that doing it twice means multiplying by $1.05^2$. A series is a sequence with plus signs between its terms, and this lesson is a shortcut for that addition.
Sequence: an ordered list of terms. Series: the sum of those terms.
Common ratio $r$: what each term is multiplied by to get the next. Divide any term by the one before it.
Finite series $S_n$: the sum of the first $n$ terms. Infinite series $S_\infty$: the sum of every term there is.
Converge: the partial sums close in on one number. $1 + \frac{1}{2} + \frac{1}{4} + \cdots$ converges to $2$; $1 + 2 + 4 + \cdots$ diverges.
Write the sum you want and the same sum multiplied by $r$, one under the other:
$$S_n = a + ar + \cdots + ar^{n-1} \qquad rS_n = ar + \cdots + ar^{n-1} + ar^n$$
Every term matches except the first of one and the last of the other, so subtracting leaves $S_n - rS_n = a - ar^n$, and
$$S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a(r^n - 1)}{r - 1}.$$
Use whichever form keeps the signs pleasant. For $2 + 6 + 18 + 54 + 162$: $a = 2$, $r = 3$, $n = 5$, so $S_5 = \frac{2(243 - 1)}{2} = 242$.
Now let $n$ grow. If $|r| < 1$ then $r^n$ shrinks towards $0$, the numerator settles on $a$, and
$$S_\infty = \frac{a}{1 - r} \quad \text{when } |r| < 1.$$
So $3 + 1 + \frac{1}{3} + \cdots = \frac{3}{1 - \frac{1}{3}} = \frac{9}{2}$. If $|r| \geq 1$ the terms never shrink and there is no sum. Money is the same series: a payment of $D$ at the end of each year for $n$ years, at rate $i$, is worth $D + D(1 + i) + \cdots + D(1 + i)^{n-1}$ when the last one is made.
Another way: picture
A bar one unit long. Shade half, then half of what is left, then half of that: the shaded part is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ and the unshaded strip halves every time. The sum is $1$, not because the adding stops but because the gap shrinks past every size you could name.
Another way: story
The formula is one trick. Multiplying the sum by $r$ slides the list along one place, so subtracting cancels everything in the middle and leaves two terms.
Using the term formula for the sum. $ar^{n-1}$ is the $n$th term; $\frac{a(r^n - 1)}{r - 1}$ is the sum. For $2, 6, 18, 54, 162$ the fifth term is $162$ and the sum is $242$.
"Adding forever must give infinity." Only if the terms stop shrinking fast enough. $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ never passes $1$: each new term covers half of what is left.
Finding $r$ by subtracting. In $3, 6, 12$ the ratio is $6 \div 3 = 2$, not $6 - 3 = 3$. Differences belong to arithmetic sequences, ratios to geometric ones.
$a = 5$ and $r = 10 \div 5 = 2$, and we want $n = 6$.
Read $a$ and $r$ off the sequence before touching the formula.
$S_6 = \frac{5(2^6 - 1)}{2 - 1} = \frac{5 \times 63}{1}$.
$2^6 = 64$, so the numerator bracket is $63$.
$S_6 = 315$.
Check: $5 + 10 + 20 + 40 + 80 + 160 = 315$.
$r = 4 \div 12 = \frac{1}{3}$, and $|r| < 1$, so there is a sum.
Check the ratio before using the infinite formula.
$S_\infty = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}}$.
Dividing by a fraction multiplies by its reciprocal.
$12 \times \frac{3}{2} = 18$.
The ratio is $\frac{1}{4}$, smaller than $1$ in size, so $S_\infty = \frac{a}{1 - r}$ applies with $a = 1$.
$1 - \frac{1}{4} = \frac{3}{4}$, so the sum is $1 \div \frac{3}{4} = \frac{4}{3}$.
A sequence starts $3, 12, 48, \ldots$, each term $4$ times the one before. What is the sum of the first $5$ terms?
Answer:
$36 + \frac{36}{5} + \frac{36}{25} + \cdots$ — what does this add up to?
Answer:
You pay $300$ into an account at the end of each year for three years, and the account pays $30\%$ a year. How much is in it immediately after the third payment?
Answer:
An infinite geometric series has first term $3$ and common ratio $r$: $3 + 3r + 3r^2 + \cdots$. For which $r$ does it add up to a finite number?
You already work out a mean, and you know the standard deviation is a typical distance from it: two data sets can share a mean of $50$ while one huddles between $48$ and $52$ and the other sprawls from $10$ to $90$. You have read histograms and called them symmetric or skewed. This lesson takes one symmetric shape seriously enough to read percentages off it.
Normal distribution: the symmetric bell-shaped model, fixed by its mean $\mu$ and standard deviation $\sigma$.
$z$-score: how many standard deviations a value sits from the mean, $z = \frac{x - \mu}{\sigma}$. Negative below the mean, positive above.
Empirical rule: the percentages lying within one, two and three standard deviations of the mean.
Tail: the far end of the curve, beyond a cut-off.
Standardise: replace a measurement by its $z$-score, so unlike units compare.
A normal distribution is the symmetric bell behind heights, measurement errors, machine fills and the means of large samples. Two numbers fix it entirely: the mean $\mu$, under the peak, and the standard deviation $\sigma$, the width of the bell. Move $\mu$ and the curve slides; change $\sigma$ and it stretches, but its shape never changes.
So the same percentages hold for every normal distribution. This is the empirical rule: about $68\%$ of the values lie within one standard deviation of the mean, and about $95\%$ within two. A fill with $\mu = 500$ ml and $\sigma = 4$ ml therefore puts about $95\%$ of bottles between $492$ and $508$ ml.
The tails follow by symmetry: about $5\%$ lie further than two standard deviations from the mean, and the two tails share that equally, so about $2.5\%$ lie that far above and the same below. One standard deviation leaves about $16\%$ in each tail.
The $z$-score does the counting:
$$z = \frac{x - \mu}{\sigma} \qquad \text{and backwards} \qquad x = \mu + z\sigma.$$
A mark of $78$ with $\mu = 70$, $\sigma = 4$ has $z = 2$; a fill of $492$ ml with $\mu = 500$, $\sigma = 4$ has $z = -2$. As $z$-scores they are on one scale, so a mark and a millilitre can be compared and the rule read off either.
Another way: picture
A bell curve with the mean under the peak and ticks one, two and three standard deviations either side. Shade the band one deviation wide, then the band two wide: the first covers most of the hill, the second nearly all of it, leaving two thin slivers at the ends.
Another way: story
A $z$-score is a ruler whose unit is the standard deviation of whatever you are measuring. Reporting $z = 2$ says how unusual a value is without saying whether it was a height, a mark or a millilitre.
Applying the rule to any data at all. The empirical rule describes a normal distribution. Incomes are strongly skewed, and for them the percentages are simply wrong; check the shape first.
Reading a negative $z$-score as a negative value. A newborn $2$ standard deviations below a mean weight of $3.4$ kg still weighs a positive number of kilograms. The minus sign says below the mean, nothing more.
Forgetting the division. $x - \mu$ is a distance in grams, millilitres or marks. Only after dividing by $\sigma$ is it a $z$-score, and only then can a height and a fill be compared.
$182 - 170 = 12$ and $170 - 158 = 12$, and $12 = 2 \times 6$.
Both ends are the same distance out; count it in standard deviations.
So the question is "within two standard deviations of the mean".
Rewriting it that way is the whole step.
By the empirical rule that is about $95\%$.
$z = \frac{46 - 64}{9} = \frac{-18}{9} = -2$.
Negative because $46$ is below the mean.
Backwards: $x = 64 + 1 \times 9 = 73$.
One standard deviation above the mean.
So $46$ is as far below the mean as $82$ is above it.
$28 - 20 = 8 = 2 \times 4$, so $28$ months is $z = 2$.
About $95\%$ lie within two standard deviations, leaving $5\%$ outside and half of that above: roughly $2.5\%$.
A measurement is normally distributed with mean $96$ and standard deviation $10$. About what percent of the values lie between $86$ and $106$?
Answer:
A measurement has mean $53$ and standard deviation $4$. What is the $z$-score of the value $61$?
Answer:
A measurement has mean $67$ and standard deviation $11$. Which value has $z$-score $1$?
Answer:
A point sits $1$ standard deviation above the mean of a normal distribution. About what percent of the values lie above it?
Answer:
A bottling machine fills to a mean of $321$ ml with a standard deviation of $13$ ml. The line is stopped whenever a bottle is more than two standard deviations from the mean. A bottle contains $308$ ml. Stop the line?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A sequence starts $6, 12, 24, \ldots$, each term $2$ times the one before. What is the sum of the first $3$ terms?
Answer:
A measurement is normally distributed with mean $173$ and standard deviation $18$. About what percent of the values lie between $155$ and $191$?
Answer:
You can sum a geometric series, finite or infinite, and use the normal model. Sum the first five terms of $3, 6, 12, \ldots$, sum $8 + 2 + \frac{1}{2} + \cdots$ forever, and say what percent of a normal distribution lies more than two standard deviations below the mean.
8. Your turn: sum $1 + \frac{1}{4} + \frac{1}{16} + \cdots$ forever, step 2
19. Your turn: parts last $\mu = 20$ months, $\sigma = 4$. What percent last past $28$ months?, step 2