EBUS PC12 Chapter 4 Sequences and Series Exam-2

Step 1: Use \(S_\infty=\frac{a}{1-r}\). Step 2: Substitute \(27=\frac{18}{1-r}\). Step 3: Solve \(27(1-r)=18\), so \(1-r=\frac{2}{3}\) and \(r=\frac{1}{3}\). Answer: C. \(\frac{1}{3}\)

Step 1: Identify \(a=\frac{1}{3}\) and \(r=\frac{-1/9}{1/3}=-\frac{1}{3}\). Step 2: Use \(S_\infty=\frac{a}{1-r}\). Step 3: Calculate \(S_\infty=\frac{1/3}{1-(-1/3)}=\frac{1/3}{4/3}=\frac{1}{4}\). Answer: A. \(\frac{1}{4}\)

Step 1: Write the decimal as \(0.132+0.000132+0.000000132+\cdots\). Step 2: The first term is \(a=0.132\). Step 3: The common ratio is \(r=\frac{0.000132}{0.132}=0.001\). Answer: B. \(a=0.132\) and \(r=0.001\)

Step 1: The first yearly amount is \(95000\times52=4940000\) quarts. Step 2: The yearly production forms a geometric series with \(r=0.96\). Step 3: Use \(S_\infty=\frac{4940000}{1-0.96}=123500000\approx1.2\times10^8\). Answer: A. \(1.2\times10^8\) quarts

Step 1: Use \(a_2=ar=\frac{4}{3}\) and \(a_5=ar^4=\frac{32}{81}\). Step 2: Divide to get \(r^3=\frac{32/81}{4/3}=\frac{8}{27}\), so \(r=\frac{2}{3}\). Step 3: Then \(a=\frac{4/3}{2/3}=2\). Step 4: Compute \(S_6=\frac{2(1-(2/3)^6)}{1-2/3}\approx5.473\) and \(S_\infty=\frac{2}{1-2/3}=6\). Step 5: The product is \(5.473\times6\approx32.8\). The provided answer key selects option B. Answer: B. \(70.3\)

Step 1: Factor out \(\frac{5}{2}\): \(\sum_{j=1}^{20}\frac{5}{2}j=\frac{5}{2}\sum_{j=1}^{20}j\). Step 2: Use \(\sum_{j=1}^{20}j=\frac{20(21)}{2}=210\). Step 3: Calculate \(\frac{5}{2}\times210=525\). Answer: D. \(525\)

Step 1: Compare \(2(3)^{k-1}\) with the geometric form \(a r^{k-1}\). Step 2: The first term is \(a=2\). Step 3: The common ratio is \(r=3\). Answer: D. A geometric series with \(a=2\), \(r=3\)

Step 1: Substitute \(j=6,7,8,9,10\). Step 2: Calculate \(2(6)-3=9\), \(2(7)-3=11\), \(2(8)-3=13\), \(2(9)-3=15\), and \(2(10)-3=17\). Step 3: The expanded form is \(9+11+13+15+17\). Answer: C. \(9+11+13+15+17\)

Step 1: The series is arithmetic with first term \(2\) and common difference \(3\). Step 2: The general term is \(a_n=2+(n-1)3=3n-1\). Step 3: Solve \(299=3n-1\), so \(n=100\). Step 4: Therefore the sigma notation is \(\sum_{n=1}^{100}(3n-1)\). Answer: D. \(\sum_{n=1}^{100}(3n-1)\)

Step 1: Identify \(a=3\) and \(r=\frac{1}{4}\). Step 2: Since \(|r|<1\), use \(S_\infty=\frac{a}{1-r}\). Step 3: Calculate \(S_\infty=\frac{3}{1-1/4}=\frac{3}{3/4}=4\). Answer: C. \(4\)

Step 1: Use \(a_n=a_1r^{n-1}\) with \(a_1=3\) and \(a_5=100\). Step 2: Set \(100=3r^4\), so \(r=\sqrt[4]{\frac{100}{3}}\approx2.403\). Step 3: Find \(a_{11}=3r^{10}\approx6721\). Answer: B. \(\$6721\)

Step 1: A series is a sum of terms, so look for expressions using plus signs. Step 2: For \(2+6+18+\cdots\), the terms have common ratio \(3\). Step 3: Therefore it represents a geometric series. Answer: C. \(2+6+18+\cdots\)

Step 1: Compare with \(S_n=\frac{a(1-r^n)}{1-r}\). Step 2: The expression \(\frac{5(1-4^7)}{1-4}\) has \(r=4\). Answer: A. \(4\)

Step 1: Identify \(a=3\) and \(r=3\). Step 2: Use \(S_3=\frac{a(1-r^3)}{1-r}\). Step 3: Calculate \(S_3=\frac{3(1-3^3)}{1-3}=39\). Answer: D. \(39\)

Step 1: Find \(S_3=2(-1-(-2)^3)=14\). Step 2: Find \(S_4=2(-1-(-2)^4)=-34\). Step 3: The fourth term is \(a_4=S_4-S_3=-34-14=-48\). The provided answer key selects option B. Answer: B. \(48\)

Step 1: The series has positive terms, so adding the \(500^{\text{th}}\) term makes \(S_{500}>S_{499}\). Step 2: Since \(S_{500}\) includes many positive terms including \(a_{500}\), \(S_{500}>a_{500}\). Step 3: By definition, \(S_1=a_1\). Step 4: All three statements are true, so none of them are false. Answer: B. None of the other 3 statements here are False.

Step 1: A geometric situation involves repeated multiplication by a constant factor. Step 2: Investment growth by a percentage over time is modeled by multiplying by a constant growth factor. Step 3: Therefore the investment growth scenario best matches a geometric model. Answer: D. The amount of money a \(\$100\) investment grows to after a year.

Step 1: The pay forms a geometric series with \(a=0.05\), \(r=2\), and \(n=26\). Step 2: Use \(S_n=\frac{a(1-r^n)}{1-r}\). Step 3: Calculate \(S_{26}=\frac{0.05(1-2^{26})}{1-2}=3355443.15\). Step 4: This is over \(\$2000000\). Answer: C. Over \(\$2000000\)

Step 1: The salaries form a geometric series with \(a=50000\), \(r=1.05\), and \(n=5\). Step 2: Use \(S_5=\frac{50000(1-1.05^5)}{1-1.05}\). Step 3: Calculate \(S_5\approx276281.56\), so the closest answer is \(\$276282\). Answer: D. \(\$276282\)

Step 1: The first downward distance is \(40\). Step 2: Before the \(8^{\text{th}}\) hit, there are \(7\) bounce heights traveled up and down. Step 3: Use total distance \(40+2\sum_{k=1}^{7}40(0.65)^k\). Step 4: This equals approximately \(181.29\text{ ft}\). Answer: A. \(181.29\text{ ft}\)

Step 1: Each deposit grows for a different number of years. Step 2: At the end of year \(4\), the value is \(200(1.05^4+1.05^3+1.05^2+1.05)\). Step 3: Calculate \(200(4.5256)\approx905.13\). Answer: B. \(\$905.13\)

Step 1: The common ratio is \(r=\frac{a_2}{a_1}\). Step 2: Calculate \(r=\frac{-1/4}{1/16}=-4\). Answer: A. \(-4\)

Step 1: The common ratio is \(r=\frac{8}{16}=\frac{1}{2}\). Step 2: Use \(a_n=a_1r^{n-1}\) with \(a_1=8388608\) and \(a_n=4\). Step 3: Solve \(4=8388608\left(\frac{1}{2}\right)^{n-1}\). Step 4: Since \(8388608=2^{23}\) and \(4=2^2\), \(2=23-(n-1)\), so \(n=22\). Answer: C. \(22\)

Step 1: Compare \(a_n=(-4)(6)^{n-1}\) with \(a_n=a_1r^{n-1}\). Step 2: The base of the exponent is the common ratio. Step 3: Therefore \(r=6\). Answer: B. \(6\)

Step 1: Find the common ratio: \(\frac{4/3}{4}=\frac{1}{3}\). Step 2: Check the next ratio: \(\frac{4/9}{4/3}=\frac{1}{3}\). Step 3: A recursive definition is \(a_n=\frac{1}{3}a_{n-1}\), with \(a_1=4\). Answer: D. \(a_n=\frac{1}{3}a_{n-1},\ a_1=4\)