SCIDES PC12 U1-Quiz 1-2 (Lesson 2,3,4)

Step 1. A sequence is arithmetic if the difference between consecutive terms is constant. Step 2. Check each sequence. For \(-11,-7,-3,1\): \[ -7-(-11)=4,\quad -3-(-7)=4,\quad 1-(-3)=4 \] This is arithmetic. For \(0.75,4,7.25,10.5\): \[ 4-0.75=3.25,\quad 7.25-4=3.25,\quad 10.5-7.25=3.25 \] This is arithmetic. For \(8,-6,4,-2\): \[ -6-8=-14,\quad 4-(-6)=10,\quad -2-4=-6 \] The differences are not constant, so this is not arithmetic. For \(3,6,9,12\): \[ 6-3=3,\quad 9-6=3,\quad 12-9=3 \] This is arithmetic. Answer: C

Step 1. An arithmetic series has a constant common difference. Step 2. Check each series. For \(-3+7+18+30+\cdots\): \[ 7-(-3)=10,\quad 18-7=11,\quad 30-18=12 \] The common difference is not constant, so this is not arithmetic. For \((-8)+(-2)+4+10+\cdots\): \[ (-2)-(-8)=6,\quad 4-(-2)=6,\quad 10-4=6 \] This is arithmetic. For \(57+69+81+93+\cdots\): \[ 69-57=12,\quad 81-69=12,\quad 93-81=12 \] This is arithmetic. For \(4+9+14+19+\cdots\): \[ 9-4=5,\quad 14-9=5,\quad 19-14=5 \] This is arithmetic. Therefore, \(-3+7+18+30+\cdots\) is not an arithmetic series. Answer: A

Step 1. Identify the first term, common difference, and number of terms. \[ a=3,\quad d=7-3=4,\quad n=10 \] Step 2. Use the arithmetic series sum formula. \[ S_n=\frac{n}{2}\left(2a+(n-1)d\right) \] Step 3. Substitute the values. \[ S_{10}=\frac{10}{2}\left(2(3)+(10-1)(4)\right) \] Step 4. Simplify. \[ \begin{aligned} S_{10}&=5(6+36)\\&=5(42)\\&=210\end{aligned} \] Answer: B

Step 1. Use the sequence equation. \[ t_n=-0.5n+5 \] Step 2. Make a table of values. \[ \begin{array}{c|ccccccccccc} n&0&1&2&3&4&5&6&7&8&9&10\\\hlinet_n&5&4.5&4&3.5&3&2.5&2&1.5&1&0.5&0\end{array} \] Step 3. The graph should start at \((0,5)\) and decrease by \(0.5\) each time \(n\) increases by \(1\). Step 4. Graph B matches these points. Answer: B

Step 1. Identify the first term, common difference, and last term. \[ a=11,\quad d=15-11=4,\quad t_n=435 \] Step 2. Find the number of terms using the general term formula. \[ t_n=a+(n-1)d \] Step 3. Substitute the values. \[ 435=11+(n-1)(4) \] Step 4. Solve for \(n\). \[ \begin{aligned} 435&=11+4n-4\\435&=4n+7\\428&=4n\\107&=n\end{aligned} \] Step 5. Use the arithmetic series sum formula with first and last terms. \[ S_n=\frac{n(a+l)}{2} \] Step 6. Substitute \(n=107\), \(a=11\), and \(l=435\). \[ \begin{aligned} S_{107}&=\frac{107(11+435)}{2}\\&=\frac{107(446)}{2}\\&=107(223)\\&=23861\end{aligned} \] Answer: D