AP Precalculus -2.1 Arithmetic and Geometric Sequences- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -2.1 Arithmetic and Geometric Sequences- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -2.1 Arithmetic and Geometric Sequences- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
Question
(B) \( g_n = 8(2)^{(n-1)} \)
(C) \( g_n = 8 \left( \frac{1}{2} \right)^n \)
(D) \( g_n = 16 \left( \frac{1}{2} \right)^{(n-1)} \)
▶️ Answer/Explanation
From the graph, likely \( g_1 = 8 \), \( g_2 = 4 \), so common ratio \( r = \frac{4}{8} = \frac{1}{2} \).
General form: \( g_n = g_1 \cdot r^{\,n-1} = 8 \cdot \left( \frac{1}{2} \right)^{n-1} \).
But none match exactly — check options:
(A) \( 4 \left( \frac{1}{2} \right)^{n-2} = 4 \cdot \left( \frac{1}{2} \right)^{n-2} = 8 \cdot \left( \frac{1}{2} \right)^{n-1} \), yes, because \( 4 \cdot \left( \frac{1}{2} \right)^{-1} = 4 \cdot 2 = 8 \).
So (A) is equivalent to the standard form.
✅ Answer: (A)
Question
(B) The exponential function \( C(x) = s \cdot p^x \) for \( 1 \leq x \leq 8 \)
(C) The arithmetic sequence \( C_n = s + pn \) for \( 1 \leq n \leq 8 \)
(D) The geometric sequence \( C_n = s \cdot p^n \) for \( 1 \leq n \leq 8 \)
▶️ Answer/Explanation
Possible costs depend on the number of plants \( n \), where \( n \) is an integer from 1 to 8.
Total cost = shovel cost + \( n \) × plant cost = \( s + pn \).
This describes a set of discrete values (not a continuous function) because \( n \) is an integer.
An arithmetic sequence with first term \( s+p \) and common difference \( p \) models this exactly: \( C_n = s + pn \), \( n = 1,2,\dots,8 \).
✅ Answer: (C)
Question
(B) The linear function \( f(n) = 2 + 4(n – 1) \)
(C) The geometric sequence \( g_n = 2^{(n-1)} \)
(D) The exponential function \( h(n) = 2 \cdot 2^{(n-1)} \)
▶️ Answer/Explanation
Test each option with inputs \( n = 2 \) and \( n = 3 \):
(A) \( a_2 = 4 \cdot 2 = 8 \) ❌ (needs 4)
(B) \( f(2) = 2 + 4(2-1) = 6 \) ❌ (needs 4)
(C) \( g_2 = 2^{1} = 2 \) ❌ (needs 4)
(D) \( h(2) = 2 \cdot 2^{1} = 4 \), \( h(3) = 2 \cdot 2^{2} = 8 \) ✅
So \( h(n) = 2 \cdot 2^{(n-1)} \) matches both points.
✅ Answer: (D)
Question
(B) If the sequence is arithmetic, the third term could be \( 12 \).
(C) If the sequence is geometric, the fifth term could be \( 48 \).
(D) If the sequence is arithmetic, the sixth term could be \( 48 \).
▶️ Answer/Explanation
Let’s analyze both possibilities:
Geometric: Let \( a_2 = ar = 6 \), \( a_4 = ar^3 = 24 \). Dividing: \( r^2 = 4 \) so \( r = 2 \) or \( r = -2 \).
If \( r = 2 \), \( a = 3 \), terms: 3, 6, 12, 24, 48 → (C) says 5th term could be 48, which is true.
If \( r = -2 \), \( a = -3 \), terms: -3, 6, -12, 24, -48 → 5th term is -48, but “could be” allows for the positive case.
Arithmetic: Let \( a_2 = a + d = 6 \), \( a_4 = a + 3d = 24 \). Subtract: \( 2d = 18 \) so \( d = 9 \), \( a = -3 \). Terms: -3, 6, 15, 24, 33, 42 → (D) says 6th term could be 48, which is false.
Check (A): If geometric and first term 1, then \( r = 6 \) so 4th term \( 1 \cdot r^3 = 216 \), not 24 ❌
Check (B): Arithmetic third term: \( a_3 = a_2 + d = 6 + 9 = 15 \), not 12 ❌
✅ Answer: (C)
Question
(B) \( 17 + 3n \)
(C) \( 21 + 3n \)
(D) \( 51 \cdot 3^{(n-10)} \)
▶️ Answer/Explanation
Simplify \( a_n = 51 + 3(n – 10) \):
\( a_n = 51 + 3n – 30 = 21 + 3n \).
This matches option (C).
Check initial term \( a_0 = 21 \) (since \( a_0 \) is given as the initial value in the problem statement).
✅ Answer: (C)
