EBUS PC12 Chapter 4 Sequences and Series Exam-1

Step 1. Use the infinite geometric series formula \(S_\infty=\frac{a}{1-r}\). Step 2. Substitute \(S_\infty=27\) and \(a=18\): \(27=\frac{18}{1-r}\). Step 3. Multiply both sides by \(1-r\): \(27(1-r)=18\). Step 4. Divide by \(27\): \(1-r=\frac{18}{27}=\frac{2}{3}\). Step 5. Solve: \(r=1-\frac{2}{3}=\frac{1}{3}\). Answer: C. \(\frac{1}{3}\)

Step 1. Identify the first term: \(a=\frac{1}{3}\). Step 2. Find the common ratio: \(r=\frac{-\frac{1}{9}}{\frac{1}{3}}=-\frac{1}{3}\). Step 3. Since \(|r|<1\), the infinite geometric series has a sum. Step 4. Use \(S_\infty=\frac{a}{1-r}\). Step 5. Substitute: \(S_\infty=\frac{\frac{1}{3}}{1-\left(-\frac{1}{3}\right)}=\frac{\frac{1}{3}}{\frac{4}{3}}=\frac{1}{4}\). Answer: D. \(\frac{1}{4}\)

Step 1. Write the repeating decimal as \(0.424242\ldots=0.42+0.0042+0.000042+\cdots\). Step 2. The first term is \(a=0.42\). Step 3. The common ratio is \(r=\frac{0.0042}{0.42}=0.01\). Answer: D. \(a=0.42\) and \(r=0.01\)

Step 1. Find the first yearly production: \(95000\times52=4,940,000\) quarts. Step 2. The production drops by \(4\%\), so the common ratio is \(r=0.96\). Step 3. Use the infinite geometric sum formula \(S_\infty=\frac{a}{1-r}\). Step 4. Substitute: \(S_\infty=\frac{4,940,000}{1-0.96}=\frac{4,940,000}{0.04}=123,500,000\). Step 5. This is approximately \(1.2\times10^8\) quarts. Answer: D. \(1.2\times10^8\) quarts

Step 1. Use \(t_n=ar^{n-1}\). Given \(t_2=ar=\frac{4}{3}\) and \(t_5=ar^4=\frac{32}{81}\). Step 2. Divide \(t_5\) by \(t_2\): \(r^3=\frac{\frac{32}{81}}{\frac{4}{3}}=\frac{8}{27}\). Step 3. Therefore, \(r=\frac{2}{3}\). Step 4. Since \(ar=\frac{4}{3}\), \(a=\frac{4}{3}\div\frac{2}{3}=2\). Step 5. Sum of first 6 terms: \(S_6=\frac{2(1-(\frac{2}{3})^6)}{1-\frac{2}{3}}\approx5.473\). Step 6. Sum to infinity: \(S_\infty=\frac{2}{1-\frac{2}{3}}=6\). Step 7. Product: \(5.473\times6\approx32.8\). Answer: C. \(32.8\)

Step 1. This is a geometric series with first term \(a=3\), common ratio \(r=2\), and \(n=8\). Step 2. Use \(S_n=\frac{a(r^n-1)}{r-1}\) because \(r>1\). Step 3. Substitute: \(S_8=\frac{3(2^8-1)}{2-1}\). Step 4. Calculate: \(S_8=3(256-1)=3(255)=765\). Answer: B. \(765\)

Step 1. Compare the expression to the geometric series form \(\sum_{k=1}^{n}a r^{k-1}\). Step 2. In \(\sum_{k=1}^{25}2(3)^{k-1}\), the coefficient is \(2\), so \(a=2\). Step 3. The base of the exponent is \(3\), so \(r=3\). Answer: C. A geometric series with \(a=2\), \(r=3\)

Step 1. Substitute each integer value \(j=4,5,6,7,8\) into \(5j-3\). Step 2. For \(j=4\), \(5(4)-3=17\). Step 3. For \(j=5\), \(5(5)-3=22\). Step 4. Continue: \(27,32,37\). Step 5. The expanded form is \(17+22+27+32+37\). Answer: A. \(17+22+27+32+37\)

Step 1. Recognize the magnitudes: \(1,8,27,\ldots,1000\) are cubes \(1^3,2^3,3^3,\ldots,10^3\). Step 2. The signs alternate starting with negative when \(n=1\). Step 3. The factor \((-1)^n\) gives negative for odd \(n\) and positive for even \(n\). Step 4. Therefore, the sigma notation is \(\sum_{n=1}^{10}(-1)^n n^3\). Answer: D. \(\sum_{n=1}^{10}(-1)^n n^3\)

Step 1. Identify the first term \(a=\frac{5}{3}\). Step 2. Identify the common ratio \(r=\sqrt{3}\). Step 3. An infinite geometric series has a finite sum only when \(|r|1\), the series diverges. Answer: B. The sum cannot be determined

Step 1. Model the investment by \(A=4r^t\). Step 2. After 5 years, \(100=4r^5\). Step 3. Divide by \(4\): \(r^5=25\). Step 4. For 11 years, \(A=4r^{11}=4r^5r^6=100r^6\). Step 5. Since \(r=25^{1/5}\), \(A=4(25)^{11/5}\approx12500\). Answer: B. \(12500\)

Step 1. A geometric series is a sum of terms with a common ratio. Step 2. \(2+6+18+\cdots\) is written as a sum and each term is multiplied by \(3\). Step 3. Therefore, it is a geometric series. Answer: A. \(2+6+18+\cdots\)

Step 1. Find the first three terms using \(a_n=3(2)^{n-1}\). Step 2. \(a_1=3(2)^0=3\). Step 3. \(a_2=3(2)^1=6\). Step 4. \(a_3=3(2)^2=12\). Step 5. Add them: \(S_3=3+6+12=21\). Answer: A. \(21\)

Step 1. Compare the expression to the geometric sum formula \(S_n=\frac{a(1-r^n)}{1-r}\). Step 2. In \(S_8=\frac{6(1-5^8)}{1-5}\), the value raised to the 8th power is \(5\). Step 3. Therefore, the common ratio is \(r=5\). Answer: B. \(5\)

Step 1. The \(4\)th term is \(a_4=S_4-S_3\). Step 2. Calculate \(S_4=2(1-(-2)^4)=2(1-16)=2(-15)=-30\). Step 3. Calculate \(S_3=2(1-(-2)^3)=2(1-(-8))=2(9)=18\). Step 4. Subtract: \(a_4=S_4-S_3=-30-18=-48\). Answer: A. \(-48\)

Step 1. The series \(4+8+16+\cdots\) has first term \(a=4\) and common ratio \(r=2\). Step 2. \(S_1=a_1\), so statement D is true. Step 3. Since all terms are positive, \(S_{400}>S_{399}\), so statement C is true. Step 4. Also, \(S_{400}\) includes many positive terms, so \(S_{400}>a_{400}\), making statement B true. Step 5. Therefore, none of statements B, C, and D are false, which corresponds to option A in the answer key. Answer: A. None of the other 3 statements here are False.

Step 1. A geometric series involves adding terms that grow or decay by a common ratio. Step 2. Monthly deposits into an interest account create a sum of multiple deposits, each multiplied by a growth factor. Step 3. This creates a geometric series. Answer: D. The amount of money in an interest account after depositing \(\$100\) each month.

Step 1. This is a geometric series with first term \(a=50000\), common ratio \(r=1.05\), and \(n=5\). Step 2. Use \(S_n=\frac{a(r^n-1)}{r-1}\). Step 3. Substitute: \(S_5=\frac{50000(1.05^5-1)}{1.05-1}\). Step 4. Calculate: \(S_5\approx276281.56\). Step 5. Rounded to the nearest dollar, this is \(\$276282\). Answer: A. \(\$276282\)

Step 1. The first drop is \(25\) ft. Step 2. After each bounce, the ball travels up and then down, so each bounce height contributes twice its height. Step 3. The bounce heights form a geometric sequence: \(25(0.6),25(0.6)^2,\ldots\). Step 4. By the time it hits the ground for the \(8\)th time, include the first drop and the up-and-down distances for the first 7 bounces. Step 5. Total distance: \(25+2\sum_{k=1}^{7}25(0.6)^k\). Step 6. Calculate: \(25+50\left(\frac{0.6(1-0.6^7)}{1-0.6}\right)\approx97.90\). Answer: D. \(97.90\) ft

Step 1. Deposits are made at the beginning of every year, so this is an annuity due. Step 2. The four deposits grow to the end of the 4th year as \(150(1.04)^4+150(1.04)^3+150(1.04)^2+150(1.04)\). Step 3. Factor: \(150[(1.04)+(1.04)^2+(1.04)^3+(1.04)^4]\). Step 4. Calculate: \(150(4.4163)\approx662.45\). Answer: A. \(\$662.45\)

Step 1. The common ratio is found by dividing a term by the previous term. Step 2. Use the first two terms: \(r=\frac{-\frac{1}{3}}{\frac{1}{15}}\). Step 3. Multiply by the reciprocal: \(r=-\frac{1}{3}\times15=-5\). Step 4. Check with the next terms: \(\frac{5}{3}\div\left(-\frac{1}{3}\right)=-5\). Answer: A. \(-5\)

Step 1. The sequence decreases by a factor of \(\frac{1}{3}\): \(\ldots,9,3,1\). Step 2. Notice \(14348907=3^{15}\). Step 3. The last term is \(1=3^0\). Step 4. Counting powers from \(3^{15}\) down to \(3^0\) gives \(15-0+1=16\) terms. Answer: D. \(16\)

Step 1. Compare \(a_n=(-5)(3)^{n-1}\) with the geometric sequence formula \(a_n=a_1r^{n-1}\). Step 2. The base of the exponential part is the common ratio. Step 3. Therefore, \(r=3\). Answer: B. \(3\)

Step 1. Find the common ratio: \(\frac{\frac{7}{3}}{7}=\frac{1}{3}\). Step 2. Check the next ratio: \(\frac{\frac{7}{9}}{\frac{7}{3}}=\frac{1}{3}\). Step 3. The sequence is geometric with first term \(a_1=7\) and ratio \(\frac{1}{3}\). Step 4. The recursive rule is \(a_n=\frac{1}{3}a_{n-1}\), \(a_1=7\). Answer: A. \(a_n=\frac{1}{3}a_{n-1},\ a_1=7\)