Tip #40 to Get a Top SAT Math Score (page 2)
The second type of sequence question has a repeating pattern, like 1, 2, 3, 1, 2, 3. … To solve this type, write out terms until you see the repeating pattern. Then use the pattern to answer the question.
Let's look at this question:
Solution: Follow the instructions for the sequence that are given in the question, until you notice a repeating pattern. Then use this repeating pattern to answer the question.
- 5, 9, –9, –5, 5, 9, –9, –5, 5
Bingo, there's the pattern. 5, 9, –9, –5, …
So use this pattern to count to the 26th term, which will be 9.
Correct answer: D
Sometimes the pattern does not repeat, but we can determine the amount that it changes between each term and use that to predict a later term. For example, in the pattern 3, 7, 11, 15, 19, …, each term after the first increases by 4. So if we wanted to predict the 105th term, we would add the first term, which is 3, to (104)(4), since there are 104 fours being added to reach the 105th term. 3 + (104)(4) = 419.
These questions are easy once you are familiar with the language. Let's practice.
- 3, 6, 9, 12, …
- In the sequence above, the first term is 3 and every number after that is found by adding 3 to the preceding number. What is the 77th term in the sequence?
- Jenny started a ball at the 15-centimeter mark on a long tape measure. She then rolled it forward 6 centimeters. Before it stopped rolling, the ball settled back toward her one unit. If Jenny continued this pattern for a total of 41 rolls, what mark would the ball reach?
- Jason is writing the letters of his name around the margin of his paper. After he writes the 201st letter, what will be the next letter that he writes?
- 3, 6, –6, …
- The first term in the sequence of numbers shown above is 3. The sequence above continues by adding 3 to every odd-numbered term and multiplying every even-numbered term by –1. For example, the second term is 3 + 3, and the third term is (–1) × 6. What is the 41st term in the sequence?
- A sequence of terms begins with 2x. Each additional term is found by doubling the preceding term. What is the value of the nth term?
- 3, 0, …
- The sequence above continues by subtracting 3 from every odd-numbered term and multiplying every even-numbered term by –1. For example, the second term is 3 – 3, and the third term is (–1) × 0. What is the 167th term in the sequence?
- A The sequence starts at 3 and each new term adds 3, so the 77th term will be found by starting with 3 and adding 3 for each of the next 76 terms: 3 + 76(3) = 231.
- A Draw a diagram to get the feel for this question (SAT Crasher's Rule #22: If a picture is described, draw it.) The diagram allows you to see that for each roll the ball advances 6 – 1 = 5 cm. Since she is rolling it from the 15-cm mark for 41 rolls, her final mark is 15 + 41(5) = 220.
- B Continue the sequence until you see the repeating pattern:
- 3 Continue the sequence until you see a repeating pattern:
- E Continue the pattern: 2x, 4x, 8x, 16x, 32x. This could have been a very tough question, but "Make It Real" makes it easy! We know that, for example, the 4th term equals 16x. So when n = 4, the expression should equal 16x. We try n = 4 in every answer choice and see which one yields 16x. Choices D and E work, so we need to try a second example as a tie breaker. When n = 5, the expression should equal 32x. We only need to test choices D and E, and this time, only choice E works: 25x = 32x.
- C Just continue the sequence until you see a repeating pattern:
J, A, S, O, N, J, A, S, O, N, …
Use the pattern to count to the 201st term, or since there are five members of the repeating pattern, divide 201 by 5 to see how many times it will repeat and then determine the remainder. 201/5 = 40 with 1 more to go. So, to get to the 201st term, Jason will write his name 40 times and then write just the first letter. The question asks for the term after he writes the 201st term, so the 202nd term is choice B.
3, 6, –6, –3. 3, 6, –6, –3, 3, …
You can see that the repeating pattern is 3, 6, –6, –3.
So just use the pattern to count to the 41st term, or divide 41 by 4 to see how many times you will loop the pattern. 41/4 = 10 with remainder 1. We loop the 4 terms 10 times and wind up with one more term, back at 3.
3, 0, 0, –3. 3, 0, 0, –3, 3, …
You can see that the repeating pattern is 3, 0, 0, –3.
Divide 167 by 4 to see how many times you will loop the pattern. 167/4 = 41 with remainder 3. So we loop the sequence 10 times and wind up with three more terms, at 0.
Go to: Tip #41
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