A sequence is an ordered list of numbers. The following is a sequence of odd numbers:
1, 3, 5, 7, . . .
A term of a sequence is identified by its position in the sequence. In the above sequence, 1 is the first term, 3 is the second term, etc. The ellipsis symbol (. . .) indicates that the sequence continues forever.
Example 1:
In sequence S, the 3rd term is 4, the 2nd term is three times the 1st, and the 3rd term is four times the 2nd. What is the 1st term in sequence S?
(A) 0
(B) 1/3
(C) 1
(D) 3/2
(E) 4
We know “the 3rd term of S is 4,” and that “the 3rd term is four times the 2nd.” This is equivalent to saying the 2nd term is 1/4 the 3rd term: $\frac{1}{4}$ · 4 = 1. Further, we know “the 2nd term is three times the 1st.” This is equivalent to saying the 1st term is 1/3 the 2nd term: $\frac{1}{3}$ ·1 = $\frac{1}{3}$ . Hence, the first term of the sequence is fully determined:
$\frac{1}{3}$ , 1, 4
The answer is (B).
Example 2:
Except for the first two numbers, every number in the sequence –1, 3, –3, . . . is the product of the two immediately preceding numbers. How many numbers of this sequence are odd?
(A) one
(B) two
(C) three
(D) four
(E) more than four
Since “every number in the sequence –1, 3, –3, . . . is the product of the two immediately preceding numbers,” the forth term of the sequence is –9 = 3(–3). The first 6 terms of this sequence are
–1, 3, –3, –9, 27, –243, . . .
At least six numbers in this sequence are odd: –1, 3, –3, –9, 27, –243.
The answer is (E).
An arithmetic progression is a sequence in which the difference between any two consecutive terms is the same. This is the same as saying: each term exceeds the previous term by a fixed amount. For example, 0, 6, 12, 18, . . . is an arithmetic progression in which the common difference is 6. The sequence 8, 4, 0, –4, . . . is arithmetic with a common difference of –4.
Example:
The seventh number in a sequence of numbers is 31 and each number after the first number in the sequence is 4 less than the number immediately preceding it. What is the fourth number in the sequence?
(A) 15
(B) 19
(C) 35
(D) 43
(E) 51
Since each number “in the sequence is 4 less than the number immediately preceding it,” the sixth term is 31 + 4 = 35; the fifth number in the sequence is 35 + 4 = 39; and the fourth number in the sequence is 39 + 4 = 43. The answer is (D). Following is the sequence written out:
55, 51, 47, 43, 39, 35, 31, 27, 23, 19, 15, 11, . . .
Advanced concepts: (Sequence Formulas)
Students with strong backgrounds in mathematics may prefer to solve sequence problems by using formulas. Note, none of the formulas in this section are necessary to answer questions about sequences on the GRE.
Since each term of an arithmetic progression “exceeds the previous term by a fixed amount,” we get the following:
first term | a + 0d | where a is the first term and d is the common difference |
second term | a + 1d | |
third term | a + 2d | |
fourth term | a + 3d | |
. . . . . . | ||
nth term | a + (n – 1)d | This formula generates the nth term |
The sum of the first n terms of an arithmetic sequence is
$\frac{n}{2}$ [2a + (n - 1)d]
A geometric progression is a sequence in which the ratio of any two consecutive terms is the same. Thus, each term is generated by multiplying the preceding term by a fixed number. For example, –3, 6, –12, 24, . . . is a geometric progression in which the common ratio is –2. The sequence 32, 16, 8, 4, . . . is geometric with common ratio 1/2.
Example:
What is the sixth term of the sequence 90, –30, 10, –10/3, . . . ?
(A) 1/3
(B) 0
(C) –10/27
(D) –3
(E) –100/3
Since the common ratio between any two consecutive terms is –1/3, the fifth term is $\frac{10}{9}$ = (- $\frac{1}{3}$ ) · (- $\frac{10}{3}$ ). Hence, the sixth number in the sequence is -$\frac{10}{27}$ = (- $\frac{1}{3}$ )·( $\frac{10}{9}$ ) .
The answer is (C).
Advanced concepts: (Sequence Formulas)
Note, none of the formulas in this section are necessary to answer questions about sequences on the GRE.
Since each term of a geometric progression “is generated by multiplying the preceding term by a fixed number,” we get the following:
first term | a | |
second term | ar ^{1} | where r is the common ratio |
third term | ar ^{2} | |
fourth term | ar ^{3} | |
. . . . | ||
nth term | a_{n} = ar ^{n-1} | This formula generates the nth term |
The sum of the first n terms of an geometric sequence is
$\frac{a\left(1-{r}^{n}\right)}{1-r}$