## SAT SUBJECT TEST MATH LEVEL 2

**PART 2**

**REVIEW OF MAJOR TOPICS**

**CHAPTER 3**

**Numbers and Operations**

**3.4 Sequences and Series**

**ARITHMETIC SEQUENCES**

One of the most common sequences studied at this level is an *arithmetic sequence* (or *arithmetic progression*). Each term differs from the preceding term by a common difference. The first *n* terms of an arithmetic sequence can be denoted by

*t*_{1}, *t*_{1} + *d*, *t*_{1} + 2*d, t*_{1} + 3*d*, . . . , *t*_{1} + (*n* – 1)*d*

where *d* is the common difference and *t** _{n }*=

*t*

_{1 }+ (

*n*– 1)

*d*. The sum of

*n*terms of the series constructed from an arithmetic sequence is given by the formula

or

If there is one term falling between two given terms of an arithmetic sequence, it is called their *arithmetic mean*.

**EXAMPLES**

**1. (****A****) Find the 28th term of the arithmetic sequence 2, 5, 8, . . . .**

**(****B****) Express the sum of 28 terms of the series of this sequence using sigma notation.**

**(****C****) Find the sum of the first 28 terms of the series.**

**SOLUTIONS**

**(****A****)** *t*_{n} = *t*_{1} + (*n*–1)*d**t*_{1} = 2, *d* = 3, *n* = 28*t*_{28} = 2 + 27·3 = 83

**(****B****)** or

**(****C****)**

*s*_{28} = (2 + 83) = 14 85 = 1190

**2. If t**

_{8 }**= 4 and**

*t*

_{12 }**= –2, find the first three terms of the arithmetic sequence.**

To solve these two equations for *d*, subtract the second equation from the third.

Substituting in the first equation gives . Thus,

The first three terms are

**3. In an arithmetic series, if S**

_{n }**= 3**

*n*

^{2}**+ 2**

*n*, find the first three terms.When *n* = 1, *S*_{1 }= *t*_{1}. Therefore, *t*_{1 }= 3(1)^{2 }+ 2 · 1 = 5.

Therefore, *d* = 6, which leads to a third term of 17. Thus, the first three terms are 5, 11, 17.