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TEN FOR TEN®
SEQUENCES AND REMAINDERS B
1)
The least integer in a set of consecutive integers is -8. If the sum of these integers is 19, how
many integers are in this set?
a) 8
c) 16
b) 9
d) 18
e) 19
The sum of three consecutive even integers is 186.
2)
3)
If y represents the greatest of the three integers, which of the following equations
represents the statement above?
a) 3y = 186
c) 3y – 3 = 186
b) 3y – 2 = 186
d) 3y – 6 = 186
e) 3y – 9 = 186
The first two numbers of a sequence are 7 and 9, respectively. The third number is 16, and,
in general, every number after the second is the sum of the two numbers immediately
preceding it. How many of the first 1,000 numbers in this sequence are odd?
a) 333
c) 665
b) 500
d) 666
e) 667
PLEASE READ THE ANSWERS AND EXPLANATIONS FOR PROBLEMS 1 THROUGH 3 NOW
1, 2, 1, -1, -2 …
4)
5)
6)
The first five terms of a sequence are shown above. After the second term, each term can
be obtained by subtracting from the previous term the term before that. For example, the
third term can be obtained by subtracting the first term from the second term. What is the
sum of the first 44 terms of this sequence?
a) 0
c) 7
b) 3
d) 9
e) 14
A paper chain is made by stringing Z individual paper links together in the repeating
sequence orange, white, red, blue, green, and purple. If the paper chain begins with a
orange link and is 73 links long, then the last link is
a) orange
c) blue
b) white
d) green
e) purple
[Grid In] If the sum of the consecutive integers from -23 to p, inclusive, is 75, what is the
value of p?
SEQUENCES AND REMAINDERS B
2
3, 6, 11, 18 …
7)
The first four terms of a sequence are shown above. Which of the following could be the
formula that gives the nth term of this sequence for all positive integers n?
a) 2n
c) 3n
b) 2n + 1
d) n2 + 1
e)
n2 + 2
44.404004000400004 ...
8)
The decimal number above consists of only 4’s and 0’s to the right of the decimal point.
The first 4 is followed by one 0, the second 4 is followed by two 0’s, the third 4 is followed by
three 0’s, and so on. What is the total number of 0’s between the 74th and 79th “4” in this
decimal number?
a) 380
c) 440
b) 390
d) 443
e)
459
4, 6, 12, 30, 84 …
9)
The first five terms of a sequence are shown above. Each consecutive term in the
sequence is calculated by performing the same operation on the previous term n. What is
the sixth term?
a) 133
c) 176
b) 139
d) 246
e) 252
p, r, s, t, u
10)
An arithmetic sequence is a sequence in which each term after the first is equal to the sum
of the preceding term and a constant. If the list of numbers shown above is an arithmetic
sequence, which of the following must also be an arithmetic sequence?
I.
2p, 2r, 2s, 2t, 2u
II.
p – 3, r – 3, s – 3, t – 3, u – 3
III.
p2, r2, s2, t2, u2
7/11/09
a) I only
c) III only
b) II only
d) I and II
e) II and III
TEN FOR TEN®
ANSWERS AND EXPLANATIONS
SEQUENCES AND REMAINDERS B
1)
E. Please draw a number line* whenever you’re asked about consecutive integers. As we
proceed upward on this number line, we add all of the negative integers starting with -8.
Since our addition starts in negative territory and we need to come up with a positive sum,
first let’s figure out how to get a sum of zero; we’ll need to add the integers from -8 all the
way up to … 8. OK, let’s keep adding consecutive integers: When we add the next
integer on the number line, 9, to our sum of zero, don’t we get a sum of 9 (0 + 9)? Now,
let’s add 10 to that sum of 9; when we do so, we get a sum of 19. Looking at our number
line, how many integers comprise our set? Well, we have 8 negative integers, 10 positive
integers, and zero!
2)
D. First, let’s find calculate the average (it’s 186/3, or 62). So, our three consecutive even
integers must be 60, 62, and 64. At this point, I assume that you know that when we’re
dealing with consecutive numbers, whether they’re consecutive integers or consecutive
multiples of 12, the average is also the median!! Plugging 64 (the greatest of the three
integers) in for y gives us …
3)
E. First, note that we have a three-number sequence that repeats odd, odd, even, odd,
odd, even. So, two numbers out of every three-number sequence are odd. With that in
mind, how many such three-number sequences are there in 1,000 numbers? 333.333, says
your calculator. It’s true; we can break 1,000 numbers into 333 complete three-number
sequences. Since each sequence contains two odd numbers, 333 sequences will provide
us with a total of 666 odd numbers. However, what does that annoying decimal (.333)
mean? It means that after we’ve taken out 333 sequences there is a single number left
over. Would such a number be even or odd? If you’ve been paying attention so far, you
know that any such “orphans” will always begin the next sequence. Is the first number in
each of our sequences even or odd? Right, it’s odd. So, 666 + 1 …
PLEASE RETURN AND FINISH PROBLEMS 4 THROUGH 10
4)
B. Did you read the instructions carefully? (“Each term can be obtained by subtracting
from the previous term the term before that ….”) So, to calculate, say, the 6th term, we
subtract the 4th term from the 5th term. When we extend the sequence, we find that the
sixth term is -1 [-2 – (-1)]; the seventh term is 1 [-1 – (-2)]; the eighth term is 2 [1 – (-1)]. Since
the sequence repeats beginning with its 7th term, we know that there are six terms in each
repeating set.^ Next: Since we need to come up with an overall sum, first we should
determine the sum of each six-term set. Well, 1 + 2 + 1 + (-1) + (-2) + (-1) = 0, right? So,
given that every six-term set adds up to 0, aren’t we really interested only in the sum of the
remainder when 44 is divided by 6 (7 full sets with a remainder of 2 terms)? When we’ve
removed those 7 full sets, we still have two terms left over! Wouldn’t they be the first two
terms in the next full set? So, 1 + 2 …
*
We recommend using vertical number lines, which eliminate any confusion about which negative number is “bigger.”
^
Note by how few numbers we had to extend this sequence before it repeated. If you’ve extended a sequence by
five terms and it hasn’t yet repeated, start over because you’ve done the math wrong.
SEQUENCES AND REMAINDERS B
ANSWERS AND EXPLANATIONS
2
5)
A. The six colors repeat, so we can use the same math that we would use to find the 73rd
term in a 6-term repeating sequence, right? Dividing by 6 (the number of links in the
sequence), we find that the sequence of colors will repeat 12 times with a remainder of 1
link. Wouldn’t that be the first link in a final, incomplete set?
6)
26. See the explanation for problem 1. Here, starting with -23, how far into positive territory
will we need to add to get a sum of zero? Positive 23, right? So, after we’ve added all the
numbers between negative 23 and positive 23, inclusive, we have a sum of 0. Since we’re
looking for a sum of 75, though, we need to keep adding. Adding the next consecutive
integer, 24, gives us a sum of 24 (0 + 24); adding 25 to that sum gives us 49; adding 26 gives
us 75.
7)
E. It’s important to distinguish between the value of a number and its place in a
sequence. The nth term refers to the (1st, 2nd, 3rd) term in the sequence; here, 3 is the 1st
term; 6 is the 2nd term; 11 is the 3rd. And so on. We need to identify the answer choice that
provides us with a formula we can use to relate 1st (the position in the sequence) to 3 (the
number occupying that position). Eliminate choices (a) and (d), since when we plug in 1
for n into those formulas, we get 2. Now, we’re down to (b), (c), and (e), so let’s try to
relate the 2nd position in the sequence to the number 6: eliminate choice (b), since
plugging 2 for n gives us 5, not 6; let’s try to use our remaining formulas [remember, we’re
down to (c) and (e) at this point] to relate 11: In (c) 3 times 3 (the position in the
sequence) gives us 9; so, (e) must be the answer. Let’s try it: 32 + 2 = 11; 42 + 2 = 18; what
would be the 5th term in this sequence?*
8)
A. We’re looking to add the number of zeros that follow the 74th (74), 75th (75), 76th (76),
77th (77), and 78th (78) terms. We end with the 79th term, so any zeros that follow that term
are outside the area we’ve been asked to add. (SAT directions are precise!)
9)
D. When a sequence seems to have inconsistent gaps between the terms, it’s good to
take a minute to examine those gaps to see whether the gaps make up the real
sequence! Here, the gap between 4 and 6 is 2; the gap between 6 and 12 is 6 (3 times 2);
the gap between 12 and 30 is 18 (3 times 6); the gap between 30 and 84 is 54 (3 times 18);
so, mustn’t the next gap be 3 times 54 (or 162)?
10)
D. In an arithmetic sequence, the gap between terms remains constant throughout each
set. Let’s plug in numbers 1, 2, 3, 4, and 5 (each succeeding term is equal to the previous
term plus 1). Doing so, we find that both Roman numerals I and II work fine (in Roman
numeral I, the constant is now twice as big—but it’s still a constant!); however, when we try
III, we get 1, 4, 9, 16, 25, which means that each term does not relate to the previous one
arithmetically.
*
27.
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