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Sequences and Series Practice Problems
Directions: Use scratch paper to solve the following problems.
Practice
 Each term in the following sequence is nine less than the previous term. What is the ninth term of the sequence?
 Each term in the following sequence is six more than the previous term. What is the value of x + z?
 Each term in the following sequence is more than the previous term. What is the value of a + b + c + d?
 Each term in the following sequence is –2 times the previous term. What is the seventh term of the sequence?
 Each term in the following sequence is –4 times the previous term. What is the value of xy?
 Each term in the following sequence is three times the previous term. What is the product of the 100th and 101st terms of the sequence?
 What is the 38th term of the following sequence?
101, 92, 83, 74, …
x, y, z, 7, 13, …
2, a, b, 3, c, d, 4, …
3, –6, 12, –24, …
x, y, –64, 256, …
1, 3, 9, 27, …
1, 3, 9, 27, 81, …
Solutions

The fourth term in the sequence is 74. You are looking for the ninth term, which is 5 terms after the fourth term. Because each term is nine less than the previous term, the ninth term will be 5(9) = 45 less than 74; 74 – 45 = 29. Because the number of terms is reasonable, you can check your answer by repeatedly subtracting 9: 74 – 9 = 65, 65 – 9 = 56, 56 – 9 = 47, 47 – 9 = 38, 38 – 9 = 29.

The term that follows z is 7. Because each term is 6 more than the previous term, z must be 6 less than 7. Therefore, z = 7 – 6 = 1. In the same way, y is 6 less than z, and x is 6 less than y: y = 1 – 6 = –5 and x = –5 – 6 = –11. The sum of x + z is equal to –11 + 1 = –10.

The first term in the sequence is 2. The next term in the sequence, a, is more than 2: 2. b is more than a, 2 c is more than 3: 3 . d is more than c, 3.Add the values of a, b, c, and d: 2 + 2 + 3 + 3 = 12.

The fourth term in the sequence is –24. You are looking for the seventh term, which is three terms after the fourth term. You must multiply by –2 three times, so the seventh term will be (–2)^{3} = –8 times –24. (–24)(–8) = 192. Because the number of terms is reasonable, you can check your answer by repeatedly multiplying by –2: (–24)(–2) = 48, (48)(–2) = –96, (–96)(–2) = 192.

Because each term in the sequence is –4 times the previous term, y is equal to = 16, and x = . Therefore, xy = (16)(–4) = –64.

Every term in the sequence is 3 raised to a power. The first term, 1, is 3^{0}. The second term, 3, is 3^{1}. The value of the exponent is one less than the position of the term in the sequence. The 100th term of the sequence is equal to 3^{100 – 1} = 3^{99}, and the 101st term in the sequence is equal to 3^{99} + 1 = 3^{100}. To multiply two terms with common bases, add the exponents of the terms: (3^{99})(3^{100}) = 3^{199}.

1, 3, 9, 27, 81, … is a geometric sequence. There is a constant ratio between terms: Each term is three times the previous term. You can use the formula Term n = a_{1} × r^{n – 1} to determine the nth term of this geometric sequence.
First determine the values of n, a_{1}, and r:
n = 38 (because you are looking for the 38th term)
a_{1} = 1 (because the first number in the sequence is 1)
r = 3 (because the sequence increases by a ratio of 3; each term is three times the previous term)
Now solve:
Term n = a_{1} × r^{n – 1}
Term 38 = 1 × 3^{38 – 1}
Term 38 = 1 × 3^{37}
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