SCIDES PC12 Unit 1 Test-2

Step 1. We are looking for \(n\) when \(t_n=5\frac{1}{2}\). Step 2. Identify the first term and common difference. \[ a=9\frac{1}{4},\quad d=9-9\frac{1}{4}=-\frac{1}{4} \] Step 3. Use the arithmetic sequence formula. \[ t_n=a+(n-1)d \] Step 4. Substitute the values. \[ 5\frac{1}{2}=9\frac{1}{4}+(n-1)\left(-\frac{1}{4}\right) \] Step 5. Simplify and solve. \[ \begin{aligned}5\frac{1}{2}&=9\frac{1}{4}-\frac{1}{4}n+\frac{1}{4}\\5\frac{1}{2}&=9\frac{1}{2}-\frac{1}{4}n\\5\frac{1}{2}-9\frac{1}{2}&=-\frac{1}{4}n\\-4&=-\frac{1}{4}n\\n&=16\end{aligned} \] Answer: B

Step 1. Identify the first term, common ratio, and last term. \[ a=5,\quad r=\frac{20}{5}=4,\quad a_n=20480 \] Step 2. Find the number of terms. \[ 20480=5(4^{n-1}) \] \[ 4^{n-1}=4096=4^6 \] \[ n-1=6,\quad n=7 \] Step 3. Use the geometric series formula. \[ S_n=\frac{a(1-r^n)}{1-r} \] Step 4. Substitute. \[ \begin{aligned}S_7&=\frac{5(1-4^7)}{1-4}\\&=\frac{5(1-16384)}{-3}\\&=\frac{5(-16383)}{-3}\\&=27305\end{aligned} \] Answer: B

Step 1. The ball first falls \(10\text{ m}\). Step 2. The bounce ratio is \(35\%\), so \(r=0.35\). Step 3. The first bounce height is: \[ a=10(0.35)=3.5 \] Step 4. When the ball hits the ground for the \(12\)th time, use \(n=11\) bounce heights after the initial drop. Step 5. Use the finite geometric sum formula. \[ S_{11}=\frac{3.5(1-0.35^{11})}{1-0.35} \] Step 6. The total vertical distance is the initial drop plus twice the bounce-height sum. \[ \begin{aligned}\text{Total distance}&=10+2S_{11}\\&=10+2\left(\frac{3.5(1-0.35^{11})}{1-0.35}\right)\\&\approx20.77\end{aligned} \] Answer: A

Step 1. An infinite geometric series converges only when \(|r|<1\). Step 2. Find the common ratio for each series. \[ I:\ r=\frac{-12}{24}=-\frac{1}{2} \] Since \(\left|-\frac{1}{2}\right|1\), series II diverges. \[ III:\ r=\frac{32}{64}=\frac{1}{2} \] Since \(\left|\frac{1}{2}\right|1\), series IV diverges. Therefore, the convergent series are I and III. Answer: C

Step 1. In a geometric sequence, consecutive ratios are equal. \[ \frac{m+3}{m}=\frac{m+7}{m+3} \] Step 2. Cross multiply. \[ (m+3)^2=m(m+7) \] Step 3. Expand and solve. \[ \begin{aligned}m^2+6m+9&=m^2+7m\\6m+9&=7m\\m&=9\end{aligned} \] Step 4. The first three terms are: \[ m,\ m+3,\ m+7=9,12,16 \] Answer: B

Step 1. From the graph, the first term is \(a=2\). Step 2. The common difference is: \[ d=2.75-2=0.75 \] Step 3. Use the arithmetic sequence formula. \[ t_n=a+(n-1)d \] Step 4. Substitute \(n=54\). \[ \begin{aligned}t_{54}&=2+(54-1)(0.75)\\&=2+53(0.75)\\&=2+39.75\\&=41.75\end{aligned} \] Answer: A

Step 1. Identify the first term and common ratio. \[ a=30,\quad r=\frac{-6}{30}=-\frac{1}{5} \] Step 2. Use the infinite geometric series formula. \[ S_{\infty}=\frac{a}{1-r} \] Step 3. Substitute. \[ S_{\infty}=\frac{30}{1-\left(-\frac{1}{5}\right)} \] Step 4. Simplify. \[ \begin{aligned}S_{\infty}&=\frac{30}{1+\frac{1}{5}}\\&=\frac{30}{\frac{6}{5}}\\&=25\end{aligned} \] Answer: C

Step 1. Identify the first term, common difference, and number of terms. \[ a=44,\quad d=41-44=-3,\quad n=60 \] Step 2. Use the arithmetic series formula. \[ S_n=\frac{n}{2}\left(2a+(n-1)d\right) \] Step 3. Substitute and simplify. \[ \begin{aligned}S_{60}&=\frac{60}{2}\left(2(44)+(60-1)(-3)\right)\\&=30(88-177)\\&=30(-89)\\&=-2670\end{aligned} \] Answer: D

Step 1. Identify the first term and common ratio. \[ a=-12,\quad r=\frac{-6}{-12}=\frac{1}{2} \] Step 2. Use the geometric term form \(ar^{k-1}\). \[ (-12)\left(\frac{1}{2}\right)^{k-1} \] Step 3. Find the number of terms using the last term. \[ -\frac{3}{16}=(-12)\left(\frac{1}{2}\right)^{n-1} \] \[ \frac{1}{64}=\left(\frac{1}{2}\right)^{n-1} \] \[ \left(\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^{n-1} \] \[ n=7 \] Therefore, the sigma notation is: \[ \sum_{k=1}^{7}(-12)\left(\frac{1}{2}\right)^{k-1} \] Answer: D

Step 1. An arithmetic series has a constant common difference. Step 2. Check option A. \[ -23-(-29)=6,\quad -18-(-23)=5,\quad -14-(-18)=4 \] Step 3. The differences are not constant, so option A is not an arithmetic series. Answer: A

Step 1. Look at the geometric series. \[ \sum_{i=2}^{\infty}3\left(\frac{2}{3}\right)^i \] Step 2. In a geometric series, the common ratio is the value repeatedly multiplied. Step 3. The repeated factor is \(\frac{2}{3}\). Therefore, the common ratio is \(\frac{2}{3}\). Answer: C

Step 1. Identify the first term, common difference, and last term. \[ a=6,\quad d=10-6=4,\quad t_n=50 \] Step 2. Find the number of terms. \[ \begin{aligned}50&=6+(n-1)(4)\\50&=6+4n-4\\50&=4n+2\\48&=4n\\n&=12\end{aligned} \] Step 3. Use \(S_n=\frac{n(a+l)}{2}\). \[ S_{12}=\frac{12(6+50)}{2}=336 \] Answer: B

Step 1. Use the infinite geometric series formula. \[ S_{\infty}=\frac{a}{1-r} \] Step 2. Substitute \(S_{\infty}=\frac{24}{7}\) and \(r=-\frac{3}{4}\). \[ \frac{24}{7}=\frac{a}{1-\left(-\frac{3}{4}\right)} \] Step 3. Simplify the denominator. \[ 1+\frac{3}{4}=\frac{7}{4} \] Step 4. Solve for \(a\). \[ \frac{24}{7}=\frac{a}{\frac{7}{4}} \] \[ a=\frac{24}{7}\cdot\frac{7}{4}=6 \] Answer: A

Step 1. The ball first drops \(2\text{ m}\). Step 2. The first bounce height is: \[ 2(0.50)=1 \] Step 3. The infinite sum of the upward bounce heights is: \[ S_{\text{up}}=\frac{1}{1-0.50}=2 \] Step 4. The ball travels the first drop plus each upward and downward bounce. \[ \text{Total}=2+2S_{\text{up}} \] Step 5. Substitute. \[ \text{Total}=2+2(2)=6 \] Answer: B

Step 1. Divide a term by the previous term. \[ r=\frac{t_2}{t_1} \] Step 2. Substitute the first two terms. \[ r=\frac{-6}{2}=-3 \] Step 3. Check: \(\frac{18}{-6}=-3\) and \(\frac{-54}{18}=-3\). Answer: B

Step 1. The initial height is \(a=7\), and the bounce ratio is \(r=0.57\). Step 2. After the first bounce, it is the second term. After the fifth bounce, it is the sixth term. Step 3. Use \(a_n=ar^{n-1}\). \[ a_6=7(0.57)^{6-1} \] Step 4. Calculate. \[ a_6=7(0.57)^5\approx0.4211 \] Therefore, the height is approximately \(0.42\text{ m}\). Answer: D

Step 1. Identify \(a=0.1\), \(r=\frac{0.5}{0.1}=5\), and \(n=8\). Step 2. Use the geometric sequence formula. \[ t_n=ar^{n-1} \] Step 3. Substitute. \[ \begin{aligned}t_8&=0.1(5)^{8-1}\\&=0.1(5)^7\\&=0.1(78125)\\&=7812.5\end{aligned} \] Answer: D

Step 1. Evaluate for \(k=2,3,4,5\). \[ \begin{aligned}k=2:&\quad 4(2)^{1-2}=4(2)^{-1}=2\\k=3:&\quad 4(2)^{1-3}=4(2)^{-2}=1\\k=4:&\quad 4(2)^{1-4}=4(2)^{-3}=\frac{1}{2}\\k=5:&\quad 4(2)^{1-5}=4(2)^{-4}=\frac{1}{4}\end{aligned} \] Step 2. Add the values. \[ 2+1+\frac{1}{2}+\frac{1}{4}=3\frac{3}{4} \] Note: The worked solution shows \(3\frac{3}{4}\), but the answer key marks C. This CSV keeps the original answer key as requested. Answer: C

Step 1. Use the arithmetic sequence equation. \[ t_n=4n+6 \] Step 2. Make a table of values. \[ \begin{array}{c|cccccc}n&0&1&2&3&4&5\\\hline t_n&6&10&14&18&22&26\end{array} \] Step 3. The correct graph should contain points such as \((0,6),(1,10),(2,14),(3,18),(4,22),(5,26)\). Step 4. Graph B matches this pattern. Answer: B

Step 1. Identify the first term and common ratio. \[ a=2,\quad r=\frac{12}{2}=6 \] Step 2. Use the geometric series formula. \[ S_n=\frac{a(1-r^n)}{1-r} \] Step 3. Substitute \(a=2\) and \(r=6\). \[ S_n=\frac{2(1-6^n)}{1-6} \] Step 4. Simplify the denominator. \[ S_n=\frac{2(1-6^n)}{-5} \] Answer: B