Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
 12.1 Logic 12.2 Sequences 12.3 Limits 12.4 Imaginary and Complex Numbers

 12.5 Key Formulas 12.6 Review Questions 12.7 Explanations
Explanations

1.      C

If a statement is false, the contrapositive of the statement will also be false. To find the contrapositive, you need to take the opposite of both parts of the statement and then switch the order. The contrapositive of “If it rains, it pours,” is “If it doesn’t pour, it doesn’t rain.”

2.      B

The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d, where d is the difference between the terms of an arithmetic sequence.

If the first term of a sequence is –3, and d = –3, then an = –3 –3n + 3 = –3n. So, the 30th term is –3(30) = –90.

3.      D

This sequence is an arithmetic sequence since the difference between each term is constant. The formula for the sum of the first n terms of an arithmetic sequence is:

To use this formula for this question, first calculate the values of a1 and a100 by plugging n = 1 and n = 100 into the given formula an = 6n – 3. So, we find that a1 = 6 – 3 = 3 and a100 = 600 – 3 = 597. The sum of the first 100 terms is therefore:

4.      D

To answer this question quickly and efficiently, you need to know the formula for the sum of the first n terms of a geometric sequence:

where r is the common ratio of the sequence. In this problem b1 = 32(21) = 3 and r = 2, so the formula yields the answer 3069.

5.      A

This question throws a little curveball at you because function is undefined at x = –2, since –22 – 4 = 0. However, the denominator can be factored into (x – 2)(x + 2). Then (x + 2) can be canceled from the numerator and denominator, leaving 1x–2 as the function. Evaluating this function at x = –2, you see the limit is –14.

6.      D

The powers of i repeat themselves in a cycle of four, that is in = in+4. Since i4 = 1, i5 must equal i. You can also reduce i14 by noticing that it equals i12 i2. Since 12 is a multiple of 4, i12 equals 1, so

So 3(ii14) = 3(i – (–1)) = 3(i + 1) = 3i + 3.

 Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
Test Prep Centers
SparkCollege
 College Admissions Financial Aid College Life