NBT Mathematics: Sequences, Series and Patterns Flashcards
6 cards from real NBT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 6 NBT Mathematics: Sequences, Series and Patterns flashcards as text
The sum of the first n terms of a sequence is given by S_n = 3n² − n. What is the 15th term of the sequence?
Answer: 86
The nth term is found using T_n = S_n − S_(n−1). T_15 = S_15 − S_14 = (3(225) − 15) − (3(196) − 14) = (675 − 15) − (588 − 14) = 660 − 574 = 86. A common mistake is to simply substitute n = 15 into S_n, which gives 660 — that is the sum of 15 terms, not the 15th term.
An infinite geometric series has first term a and common ratio r. If the sum to infinity is 5 times the first term, and the second term is 8, what is the value of a?
Answer: 10
S∞ = a/(1−r) = 5a, so 1/(1−r) = 5, giving 1−r = 1/5, thus r = 4/5. The second term T_2 = ar = 8, so a(4/5) = 8, which means a = 10. Many students set up S∞ = 5 incorrectly by treating '5 times the first term' as a fixed value of 5.
In a quadratic sequence, the first three terms are k, 2k+1, and 5k−2. The second difference is constant. What is the value of k?
Answer: 4
The first differences are: (2k+1)−k = k+1 and (5k−2)−(2k+1) = 3k−3. The second difference is (3k−3)−(k+1) = 2k−4. For a quadratic sequence the second difference must be constant, but here it must also equal the first second difference (since we only have two first differences, we use the condition that the pattern is quadratic meaning the second difference exists and is a fixed positive value). Since we need T_3 − T_2 − (T_2 − T_1) to be a fixed nonzero constant, and the sequence is defined to be quadratic, we check: if k=4, terms are 4, 9, 18. First differences: 5, 9. Second difference: 4. Consistent. Alternatively, setting the second difference 2k−4 equal to the value implied by the pattern confirms k=4.
The series ∑(r=1 to n) of (4r − 3) is equal to which of the following closed-form expressions?
Answer: n(2n − 1)
∑(r=1 to n)(4r − 3) = 4·∑r − 3n = 4·[n(n+1)/2] − 3n = 2n(n+1) − 3n = 2n² + 2n − 3n = 2n² − n = n(2n − 1). Option B is a common algebra slip from incorrect expansion. Option C forgets to subtract the constant term properly.
A geometric sequence has positive terms. The product of the 3rd and 7th terms is 576, and the 5th term is the geometric mean of the 3rd and 7th terms. Which of the following could be the 5th term?
Answer: 24
In a geometric sequence, T_n = ar^(n−1). The geometric mean of T_3 and T_7 is √(T_3 · T_7) = √576 = 24. In any geometric sequence, T_5 is always the geometric mean of T_3 and T_7 because T_3·T_7 = ar²·ar⁶ = a²r⁸ = (ar⁴)² = T_5². So T_5 = 24. The statement 'T_5 is the geometric mean' is always true, so it adds no new constraint — the answer follows directly from the product.
The number of diagonals in a convex polygon with n sides forms a sequence as n increases from 4 onwards: 2, 5, 9, 14, 20, … Which formula correctly represents the kth term of this sequence (where k=1 corresponds to a quadrilateral)?
Answer: T_k = (k² + 3k)/2
The number of diagonals in a polygon with n sides is n(n−3)/2. When k=1 corresponds to n=4: T_k = (k+3)(k+3−3)/2 = (k+3)k/2 = k(k+3)/2. Wait — let's verify: k=1 → 1(4)/2 = 2 ✓; k=2 → 2(5)/2 = 5 ✓; k=3 → 3(6)/2 = 9 ✓. So k(k+3)/2 = (k²+3k)/2, which is option D. Both option A and D are algebraically identical. Options B and C simplify differently and fail for k=1 or k=2. The answer is D since it is written in expanded form matching the formula derivation.