NBT Mathematics: Sequences, Series and Patterns Flashcards

In an arithmetic sequence, the 5th term is 3 and the product of the 3rd and 7th terms equals zero. Which of the following is a possible value of the common difference?

Answer: 3/2

The 3rd and 7th terms are symmetric about the 5th term: T₃ = 3 − 2d and T₇ = 3 + 2d. Their product is (3 − 2d)(3 + 2d) = 9 − 4d² = 0, giving d² = 9/4, so d = ±3/2. Of the options, 3/2 is correct.

The second term of a geometric sequence is −6 and the fifth term is 48. What is the common ratio?

Answer: −2

Dividing T₅ by T₂ eliminates a: r³ = 48 ÷ (−6) = −8. The real cube root of −8 is −2, so r = −2. Verify: a = T₂/r = −6/(−2) = 3, and T₅ = 3·(−2)⁴ = 3·16 = 48 ✓.

The sum of the first n terms of a series is given by Sₙ = 3n² − n. What is the value of the 7th term, T₇?

Answer: 38

Using Tₙ = Sₙ − Sₙ₋₁: S₇ = 3(49) − 7 = 140 and S₆ = 3(36) − 6 = 102. Therefore T₇ = 140 − 102 = 38. (Alternatively, Tₙ = 6n − 4, giving T₇ = 42 − 4 = 38.)

The infinite geometric series 1 + (x − 1) + (x − 1)² + ⋯ converges to 5. What is the value of x?

Answer: 9/5

The series has first term a = 1 and common ratio r = (x − 1). Applying S∞ = a/(1 − r): 1/(1 − (x − 1)) = 1/(2 − x) = 5, so 2 − x = 1/5, giving x = 9/5. Check convergence: |r| = |9/5 − 1| = 4/5 < 1 ✓.

A quadratic sequence has T₁ = 2, T₂ = 3, and a second constant difference of 1. What is T₁₀?

Answer: 47

For Tₙ = an² + bn + c, the second constant difference equals 2a, so 2a = 1 → a = 1/2. From T₁: 1/2 + b + c = 2, and from T₂: 2 + 2b + c = 3. Solving gives b = −1/2 and c = 2. Thus T₁₀ = (1/2)(100) + (−1/2)(10) + 2 = 50 − 5 + 2 = 47.

Evaluate: $\sum_{k=3}^{8}(3k - 1)$

Answer: 93

The six terms are: k=3→8, k=4→11, k=5→14, k=6→17, k=7→20, k=8→23. These form an arithmetic series with first term 8, last term 23, and 6 terms. Sum = 6(8 + 23)/2 = 6 × 31/2 = 93. (Note: 100 is the trap answer for summing from k=1 to 8.)