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What are arithmetic sequences and series for ?
When a ball is thrown straight up in the air, it starts out fast,
gradually slows until it reaches its maximum height, and then it
starts coming down again.
A
change
in
speed
can
be
described
as
a
change
in
distance
traveled
during
a
unit
of
time.
Each unit of time, the ball goes up less than it did during the
previous unit. Remarkably, the amount of decrease from one unit of
time to the next is constant!
The
amount
of
change
depends
on
gravity,
which
is
different
on
different
planets.
For example, suppose that the ball moves straight up 52'' in
the first tenth of a second after it is thrown. In the next 
sec, it won’t rise as much. Suppose it rises only 48''--that is,
4'' less than it did in the first  sec. Then, each tenth of a
second, it will again rise 4'' less than it did the tenth of a second
before. Here are two questions one might ask about such a
situation.
For
our
planet,
the
4''
decrease
of
upward
movement
every
tenth
of
a
second
is
very
close
to
the
true
pull
of
gravity.
- How long will it be before the ball starts coming back down?
- How high will it go?
One way to figure out answers to questions like these is to list
how far the ball travels each tenth of a second.
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| | Time after release from
hand | 1st 10th
of a sec | 2nd
10th of
a sec | 3rd 10th
of a sec | ... |
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| | Distance traveled during
that tenth of a second | 52 in. | 48 in. | 44 in. | ... |
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- Create a table like this and fill it in for each tenth of a
second for the first 16 tenths of a second.
In
this
problem,
you
are
creating
an
arithmetic
sequence,
a
sequence
of
numbers
that
differ
by
a
constant
amount
(in
this
case
-0.8)
from
term
to
term.
- What total distance does the ball travel upward during the
first 0.2 sec?
Here
you
are
summing
an
arithmetic
series.
- How far does the ball travel during the first second?
- At what time does the ball no longer go up?
Here,
you
are
finding
which
term
in
an
arithmetic
sequence
has
a
particular
value.
- From the table, how high does the ball go at its highest?
Again,
to
solve
this
you
are
summing
a
series.
- What do the negative numbers in your table mean?
- From the time that the ball has reached its highest, how
long will it be before the ball reaches your hand again?
Try
to
figure
this
out
without
extending
your
table.
Hints
Hint to problem 6. Think about where the ball is and what it is
doing just a moment earlier. Think what must happen next. Then
explain the number (both its size and its sign) in terms of what the
ball is doing.
Hint to problem 7. Compare the distances traveled in the two
tenths of a second just before and after the very top of the ball’s
trip.
Answers
See table below.
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| | Time |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
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| | Dist | 52'' | 48'' | 44'' | 40'' | 36'' | 32'' | 28'' | 24'' | 20'' | 16'' | 12'' | 8'' | 4'' | 0'' | -4'' | -8'' |
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- During the first 0.2 sec, the ball travels upward 100 inches.
- During the first 1 sec, the ball travels upward 340 inches.
- Some time during the fourteenth
 of a second the ball
doesn’t go up any more.
- By then, using the numbers above, it has gone up 364''.
- The negative numbers in the table mean that the ball is
falling back down. The amount of falling increases each
tenth of a second.
- It will take 1.4 sec.
Solutions
See table below.
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| | Time |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
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| | Dist | 52'' | 48'' | 44'' | 40'' | 36'' | 32'' | 28'' | 24'' | 20'' | 16'' | 12'' | 8'' | 4'' | 0'' | -4'' | -8'' |
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- By the end of 0.2 sec, the ball has gone 52'' + 48'' for a
total of 100''.
- By the end of 1 sec, the ball has traveled upward
52'' + 48'' + 44'' + 40'' + 36'' + 32'' + 28'' + 24'' + 20'' + 16''.
This is quite a nuisance to add up unless you notice
that
| | 52 | + 48 | + 44 | + 40 | + 36 | + 32 | + 28 | + 24 | + 20 | + 16 |
| + | 16 | + 20 | + 24 | + 28 | + 32 | + 36 | + 40 | + 44 | + 48 | + 52 |
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| | | 68 | + 68 | + 68 | + 68 | + 68 | + 68 | + 68 | + 68 | + 68 | + 68 |
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In other words, the sum of the first term and last term,
(52 + 16) = 68, times the number of terms (10), is twice
the sum that you want. So, an easy way to compute
52 + 48 + 44 + 40 + 36 + 32 + 28 + 24 + 20 + 16 is to compute
(52 + 16) × 10 ÷ 2. This is 340.
- During the fourteenth
 of a second, the ball rises 0''. This
doesn’t mean that the ball is sitting still for a tenth of a
second! It does mean that the ball’s height at the beginning
and end of that interval of time is the same (the change in
height from beginning to end is 0). In between, it goes up a bit
more and comes back down.
- Using the numbers in the table, the ball will have gone a total
of (52 + 0) × 14 ÷ 2, which is 364 inches up. In fact, making
the reasonable (and correct) assumption that the fourteenth
 -second is split evenly between “up” and “down,” the total
distance the ball travels up is 2'' more than one can read
off the table--366 inches at its highest. The way this
question is worded, either 364 or 366 is a reasonable
answer.
- The negative numbers in the table mean that the ball is falling
back down. The amount of falling increases by 4'' each tenth of
a second.
- The distance traveled each
 -second on the way down
exactly mirrors the distance traveled on the way up, so
the time it takes to get downis the same as the time
it took to get up. It took the ball 1.4 sec to reach its
highest point, so it will take another 1.4 sec to get back
down.
These
figures
come
from
the
table.
If
we
account
for
the
“extra”
travel
up
during
the
fourteenth
 -second,
it
takes
1.45
sec
to
go
up
and
the
same
to
get
down.
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