SCIDES PC12 Unit 1 Test-1

Step 1. Use the equation \(t_n=12n+6\). Step 2. Make a table of values. \[\begin{array}{c|cccccc}n&0&1&2&3&4&5\\ \hline t_n&6&18&30&42&54&66\end{array}\] Step 3. The correct graph must show points such as \((0,6),(1,18),(2,30),(3,42),(4,54),(5,66)\). Step 4. Graph A matches this pattern. Answer: A

Step 1. Identify \(a=3\), \(r=\frac{6}{3}=2\), and \(n=14\). Step 2. Use the geometric sequence formula. \[t_n=ar^{n-1}\] Step 3. Substitute. \[\begin{aligned}t_{14}&=3(2)^{14-1}\\&=3(2)^{13}\\&=3(8192)\\&=24576\end{aligned}\] Answer: B

Step 1. Identify the first term and common difference. \[a=11,\quad d=8-11=-3\] Step 2. Use the arithmetic sequence formula. \[t_n=a+d(n-1)\] Step 3. Substitute \(n=45\). \[\begin{aligned}t_{45}&=11+(-3)(45-1)\\&=11+(-3)(44)\\&=11-132\\&=-121\end{aligned}\] Answer: C

Step 1. From the graph, the first term is \(a=6\). Step 2. The values decrease by \(1\) each term, so the common difference is: \[d=5-6=-1\] Step 3. Use the arithmetic sequence formula. \[t_n=a+(n-1)d\] Step 4. Substitute \(n=11\). \[\begin{aligned}t_{11}&=6+(11-1)(-1)\\&=6+10(-1)\\&=6-10\\&=-4\end{aligned}\] Answer: C

Step 1. Identify \(a=4\), \(d=7\), and last term \(l=88\). Step 2. Find \(n\). \[\begin{aligned}88&=4+(n-1)7\\88&=7n-3\\91&=7n\\n&=13\end{aligned}\] Step 3. Use \(S_n=\frac{n(a+l)}{2}\). \[S_{13}=\frac{13(4+88)}{2}=598\] Answer: C

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

Step 1. Identify \(a=2.1\), \(d=1.9-2.1=-0.2\), and \(n=40\). Step 2. Use the arithmetic sum formula. \[S_n=\frac{n}{2}\left(2a+(n-1)d\right)\] Step 3. Substitute the values. \[\begin{aligned}S_{40}&=\frac{40}{2}\left(2(2.1)+(40-1)(-0.2)\right)\\&=20(4.2-7.8)\\&=20(-3.6)\\&=-72\end{aligned}\] Answer: B

Step 1. Identify the first term and common ratio. \[a=\frac{3}{16},\quad r=\frac{\frac{3}{8}}{\frac{3}{16}}=2\] Step 2. Use the geometric term form \(a r^{k-1}\). \[\frac{3}{16}(2)^{k-1}\] Step 3. Find the number of terms using the last term. \[1536=\frac{3}{16}(2)^{n-1}\] \[8192=2^{n-1}\] \[2^{13}=2^{n-1}\] \[n=14\] Step 4. Therefore, the sigma notation is: \[\sum_{k=1}^{14}\left(\frac{3}{16}\right)(2)^{k-1}\] Answer: D

Step 1. Evaluate the terms for \(k=2,3,4,5\). \[\begin{aligned}k=2:&\quad 4(2)^{1-2}=4(2)^{-1}=2\\k=3:&\quad 4(2)^{1-3}=4(2)^{-2}=1\\k=4:&\quad 4(2)^{1-4}=4(2)^{-3}=\frac12\\k=5:&\quad 4(2)^{1-5}=4(2)^{-4}=\frac14\end{aligned}\] Step 2. Add the terms. \[2+1+\frac12+\frac14=\frac{15}{4}=3\frac34\] Answer: C

Step 1. Use \(S_{\infty}=\frac{a}{1-r}\). Step 2. Substitute \(S_{\infty}=-\frac{243}{2}\) and \(a=-81\). \[-\frac{243}{2}=\frac{-81}{1-r}\] Step 3. Solve for \(r\). \[\begin{aligned}-243(1-r)&=2(-81)\\-243+243r&=-162\\243r&=81\\r&=\frac13\end{aligned}\] Answer: A

Step 1. Divide a term by the previous term. \[r=\frac{t_2}{t_1}\] Step 2. Substitute. \[r=\frac{-64}{-256}=\frac{1}{4}\] Answer: D

Step 1. Identify \(a=6\), \(r=\frac{-18}{6}=-3\), and \(n=9\). Step 2. Use the geometric sum formula. \[S_n=\frac{a(1-r^n)}{1-r}\] Step 3. Substitute. \[\begin{aligned}S_9&=\frac{6(1-(-3)^9)}{1-(-3)}\\&=\frac{6(1-(-19683))}{4}\\&=\frac{6(19684)}{4}\\&=29526\end{aligned}\] Answer: A

Step 1. Use \(a_n=ar^{n-1}\). \[162=2r^{5-1}\] Step 2. Solve for \(r\). \[r^4=\frac{162}{2}=81\] \[r=\pm\sqrt[4]{81}=\pm3\] Step 3. Find the second term. \[a_2=ar=2(\pm3)=\pm6\] Answer: C

Step 1. An infinite geometric series converges only when \(|r|<1\). Step 2. Check each common ratio. \[I:\ r=\frac{-12}{24}=-\frac12,\quad |r|1\] \[III:\ r=\frac{32}{64}=\frac12,\quad |r|1\] Step 3. Only I and III converge. Answer: B

Step 1. If \(1\%\) is washed out, \(99\%\) remains each time. \[r=0.99\] Step 2. After \(10\) washes, the amount remaining is: \[(0.99)^{10}\approx0.904\] Step 3. Convert to a percent. \[0.904\approx90\%\] Answer: B

Step 1. Identify \(a=2080\), \(r=\frac{2121.60}{2080}=1.02\), and \(n=52\). Step 2. Use the geometric sum formula. \[S_n=\frac{a(1-r^n)}{1-r}\] Step 3. Substitute. \[S_{52}=\frac{2080(1-1.02^{52})}{1-1.02}\] Step 4. Calculate. \[S_{52}\approx187234.13\] Answer: D

Step 1. Identify \(a=8\) and \(r=\frac{2}{8}=\frac14\). Step 2. Use the infinite geometric series formula. \[S_{\infty}=\frac{a}{1-r}\] Step 3. Substitute. \[\begin{aligned}S_{\infty}&=\frac{8}{1-\frac14}\\&=\frac{8}{\frac34}\\&=\frac{32}{3}\end{aligned}\] Answer: D

Step 1. The initial drop is \(2\text{ m}\). Step 2. The first bounce rise is: \[2(0.70)=1.4\] Step 3. The infinite sum of the bounce rises is: \[S_{\text{up}}=\frac{1.4}{1-0.70}=\frac{1.4}{0.30}=4.666\cdots\] Step 4. The ball travels the initial drop plus up-and-down bounce distances. \[\text{Total}=2+2S_{\text{up}}\] Step 5. Substitute. \[\text{Total}=2+2(4.666\cdots)=11.33\cdots\] Therefore, the total distance is about \(11.3\text{ m}\). Answer: A

Step 1. Identify \(a=2\), \(r=3\), and last term \(1458\). Step 2. Find \(n\). \[1458=2(3)^{n-1}\] \[3^{n-1}=729=3^6\] \[n=7\] Step 3. Use the geometric sum formula. \[\begin{aligned}S_7&=\frac{2(1-3^7)}{1-3}\\&=\frac{2(1-2187)}{-2}\\&=2186\end{aligned}\] Answer: B

Step 1. The first term occurs at the lower limit \(x=-2\). Step 2. Substitute \(x=-2\). \[\frac14\left(-\frac12\right)^{-2}\] Step 3. Simplify. \[\left(-\frac12\right)^{-2}=\left(-2\right)^2=4\] \[\frac14(4)=1\] Answer: B