410 McGRAW-HILL’S SAT
Concept Review 1
- A sequence is simply a list of numbers, each of
which is called a “term.” - If the sequence repeats every six terms, you can
find the 115th term by finding the remainder when
115 is divided by 6. Since 115 ÷6 equals 19 with a
remainder of 1, the 115th term will be the same as
the first term. - Begin by finding the sum of the repeating pattern.
Next, determine how many times the pattern oc-
curs in the first 32 terms: 32 ÷ 4 =8 times. Then
multiply the sum of the pattern by 8 to obtain the
sum. - Count the number of negative terms in each repeti-
tion of the pattern, then find how many times the
pattern repeats in the first 36 terms. Since 36 ÷ 5 = 7
with a remainder of 1, the pattern repeats 7 times
and is 1 term into the eighth repetition. Multiply the
number of negative terms per repetition by 7, and if
the first term of the sequence is negative, add 1 to
the total.
5. This is a geometric sequence. Each term is the
previous one times 3 (1 × 3 =3; 3 × 3 =9, etc.). The
first term of the sequence is 3^2 , and the 30th term
is 3^2 × 329 = 327.
6. The first term is 4. Second: 4(2) + 8 =16. Third:
16(2) + 8 =40. Fourth: 40(2) + 8 =88. Fifth: 88(2)- 8 =184. Sixth: 184(2) + 8 =376.
7. The pattern repeats every five terms, and each rep-
etition contains two vowels. Since 143 ÷ 5 =28 with
a remainder of 3, the first 143 letters contain 28 ×
2 =56 vowels plus the one vowel in the first three
letters of the word SCORE, for a total of 56 + 1 =57.
8. Work backwards: xwas found by subtracting 3 from
the second term and dividing by 2. Therefore, mul-
tiply xby 2 and add 3 to get the secondterm:
2 x+3. Repeat to find the first term: 2(2x+3) + 3 =
4 x+9.
9. The integers 1 through 9 represent the first 9 digits,
and 10 through 19 represent the next 20 digits. Each
integer thereafter contains 2 digits. 26 represents
the 42nd and 43rd digits, so 2 is the 44th digit.
Answer Key 1: Sequences
SAT Practice 1
- D The first term of the sequence is x.The second
term is 2(x) 3 = 2 x3. The third term is
2(2x3) 3 = 4 x 6 3 = 4 x9. The fourth term
is 2(4x9) 3 = 8 x 18 3 = 8 x21. The fifth
term is 2(8x21) 3 = 16 x 42 3 = 16 x45. - C Each term in the sequence is the previous term
times 2. The first term,^1 ⁄ 8 , is equal to 2^3. To find
the value of the 13th term, multiply the first term
by 2 twelve times or by 2^12 to get your answer.
2 ^3 × 212 = 2 ^3 +^12 = 29 - 15 The first term is 400, after which each term is
20 less than 1/2 the previous term. The second
term is^1 ⁄ 2 (400) 20 =180. The third term is^1 ⁄ 2 (180)
20 =70. The fourth term is^1 ⁄ 2 (70) 20 =15. - C The sequence contains a repeating six-term
pattern: 148285. To find out how many times the
pattern repeats in the first 500 terms, divide 500
by 6: 500 ÷ 6 = 831 ⁄ 3. By the 500th term, the pattern
has repeated 83 full times and is^1 ⁄ 3 of the way
through the 84th repetition. Each repetition of
the pattern contains two odd digits, so in the 83
full repetitions there are 83 × 2 =166 odd digits.
In the first^1 ⁄ 3 of the pattern there is one odd digit.
Therefore there are 166 + 1 = 167 odd digits.- B There will be 44 + 45 + 46 =135 6s between the
44th and 47th appearances of 5. - B In this arithmetic sequence you must add 6 to
each term. To get from the 1st to the 104th term
you will add 103 terms, or 103 6s. The value of the
104th term is thus 5 +(103)(6) =613. - B 31 =3; 3^2 =9; 3^3 =27; 3^4 =81; 3^5 =243; 3^6 =729.
The units digits repeat in the pattern 3, 9, 7, 1, 3,
9, 7, 1,.. ., and 36 ÷ 4 =nine full repetitions. Since
it goes in evenly, it must fall on the last term of the
pattern, which is 1. - 70 The pattern alternates back and forth between
210 and 70. Each odd-numbered term is 210 and each
even-numbered term is 70, so the 24th term is 70. - 5/8 or .625 The first term of the sequence is 640.
Each term thereafter is 1/4 of the immediately
preceding term. The first six terms are 640, 160,
40, 10, 2.5, .625 (.625 =5/8).