Every term in the sequence is 3 raised to a power. The first term, 1, is 30. The second term, 3, is 3¹. 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)} = 3⁹⁹ and the 101st term in the sequence is equal to 3^{(99 + 1)} = 3¹⁰⁰. To multiply two terms with common bases, add the exponents of the terms: (3⁹⁹)(3¹⁰⁰) = 3¹⁹⁹.

Since each term in the sequence is –4 times the previous term, y is equal to \(\frac{-64}{-4}\) = 16, and x =\(\frac{16}{-4}\)= –4. Therefore, xy = (16)(–4) = –64.

Every term in the sequence is 5 raised to a power. The first term,\(\frac{1}{125}\),is 5^–3.The second term,\(\frac{1}{25}\),is 5^–2.The value of the exponent is four less than the position of the term in the sequence. The 20th term of the sequence is equal to 5^{(20 – 4)} = 5¹⁶.

The fourth term in the sequence is \(\frac{16}{3}\). You are
looking for the seventh term, which is three terms after the fourth term.You must multiply by ²⁄₃ three times, so the seventh term will be (²⁄₃)³ = \(\frac{8}{27}\) times \(\frac{16}{3};\left ( \frac{8}{27} \right )\left ( \frac{16}{3} \right )=\frac{128}{81}\). Alternatively, every term in the sequence is 18 times ²⁄₃ raised to a power.The first term,18,is 18 (²⁄₃)⁰.The second term, 12, is 18 (²⁄₃)¹. The value of the exponent is one less than the position of the term in the sequence. The seventh term of the sequence is equal to \(18\times \left ( \frac{2}{3} \right )^{6}=18\times \left ( \frac{64}{729} \right )=2\times \left ( \frac{64}{81} \right )=\frac{128}{81}\).

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. Since 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.

The first term in the sequence is 2. The next term in the sequence, a, is ¹⁄₃ more than 2: 2(¹⁄₃). b is ¹⁄₃ more than a, 2(²⁄₃). c is ¹⁄₃ more than 3: 3(¹⁄₃). d is ¹⁄₃ more than c, 3(²⁄₃). Add the values of a, b, c, and d: 2(¹⁄₃) + 2(²⁄₃)+ 3(¹⁄₃) + 3(²⁄₃) = 12.

The term that follows z is 7. Since 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 term that precedes x is 5. Therefore, the value of x is 5 – 7 = –2, and the value of y is –2 – 7 = –9. Therefore, x – y = –2 – (–9) = –2 + 9 = 7.

The fourth term in the sequence is 10¹⁄₂. You are looking for the eighth term, which is four terms after the fourth term. Since each term is ³⁄₂ more than the previous term, the eighth term will be 4(³⁄₂) = 6 more than 10 ¹⁄₂; 10(¹⁄₂) + 6 = 16(¹⁄₂). Since the number of terms is reasonable, you can check your answer by repeatedly adding ³⁄₂; 10(¹⁄₂) + ³⁄₂ = 12, 12 + ³⁄₂ = 13(¹⁄₂), 13(¹⁄₂) + ³⁄₂ = 15, 15 + ³⁄₂ = 16(¹⁄₂).

The fourth term in the sequence is 74. You are looking for the ninth term, which is 5 terms after the fourth term. Since each term is nine less than the previous term, the ninth term will be 5(9) = 45 less than 74; 74 – 45 = 29. Since 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.