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AP Calculus BC 10.2 Working with Geometric Series - Exam Style Questions - MCQs - New Syllabus

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Question

The second and third terms of the geometric series \( \displaystyle \sum_{n=1}^{\infty} a_n \) are \( a_2 = -\tfrac{3}{4} \) and \( a_3 = \tfrac{1}{4} \), respectively. What is the sum of the series \( \displaystyle \sum_{n=1}^{\infty} a_n \)?
A) \( -\tfrac{9}{16} \)
B) \( \tfrac{1}{16} \)
C) \( \tfrac{27}{16} \)
D) \( \tfrac{27}{8} \)
▶️ Answer/Explanation
Detailed solution

Ratio: \( r=\dfrac{a_3}{a_2}=\dfrac{\tfrac{1}{4}}{-\tfrac{3}{4}}=-\tfrac{1}{3} \).
First term: \( a_1=\dfrac{a_2}{r}=\dfrac{-\tfrac{3}{4}}{-\tfrac{1}{3}}=\tfrac{9}{4} \).
Sum of an infinite geometric series (\(|r|<1\)): \( S=\dfrac{a_1}{1-r}=\dfrac{\tfrac{9}{4}}{1-(-\tfrac{1}{3})} =\dfrac{\tfrac{9}{4}}{\tfrac{4}{3}} =\tfrac{27}{16}. \)

✅ Correct: C) \( \tfrac{27}{16} \)

Question

What are all values of \(x\) for which the series \[ \sum_{n=0}^{\infty} (-1)^n (3x+1)^n \] converges?
(A) \(-\dfrac{1}{3} \le x < \dfrac{1}{3}\)
(B) \(-\dfrac{1}{3} < x \le \dfrac{1}{3}\)
(C) \(-\dfrac{2}{3} \le x < 0\)
(D) \(-\dfrac{2}{3} < x < 0\)

▶️ Answer/Explanation
Detailed solution

Recognize a geometric series: \[ \sum_{n=0}^{\infty} (-1)^n (3x+1)^n = \sum_{n=0}^{\infty} \big(-3x-1\big)^n, \] with ratio \(r=-3x-1\).
A geometric series converges when \(|r|<1\). Hence \[ |-3x-1|<1 \iff |3x+1|<1 \iff -1<3x+1<1 \iff -\tfrac{2}{3}<x<0. \] Endpoints: \(x=-\tfrac{2}{3}\Rightarrow r=1\) (diverges); \(x=0\Rightarrow r=-1\) (diverges).
✅ Answer: (D) \(-\dfrac{2}{3}<x<0\)

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