An arithmetic series is the sum of terms from an arithmetic sequence, where each term changes by the same common difference. These problems practice finding sums from sigma notation, using the arithmetic series formulas, and solving for missing values like \(a_1\), \(d\), \(a_n\), and \(S_n\). Arithmetic series problems often connect the explicit formula \(a_n=a_1+(n-1)d\) with the sum formulas \(S_n=\frac{n}{2}(a_1+a_n)\) and \(S_n=\frac{n}{2}(2a_1+(n-1)d)\).
Notes
Practice Problems
\(\textbf{1)}\) \(\displaystyle\sum_{i=3}^{5}3-2i\) The answer is \(-15\)
\(\,\,\,\,\,\displaystyle\sum_{i=3}^{5}3-2i\)
\(\,\,\,\,\,=(3-2(3))+(3-2(4))+(3-2(5))\)
\(\,\,\,\,\,=(3-6)+(3-8)+(3-10)\)
\(\,\,\,\,\,=-3+(-5)+(-7)\)
\(\,\,\,\,\,=-15\)
\(\,\,\,\,\,\displaystyle\sum_{i=3}^{5}3-2i\)
\(\,\,\,\,\,=(3-2(3))+(3-2(4))+(3-2(5))\)
\(\,\,\,\,\,=(3-6)+(3-8)+(3-10)\)
\(\,\,\,\,\,=-3+(-5)+(-7)\)
\(\,\,\,\,\,=-15\)
\(\textbf{2)}\) \(\displaystyle\sum_{i=4}^{9}3i-5\) The answer is \(87\)
\(\,\,\,\,\,\displaystyle\sum_{i=4}^{9}3i-5\)
\(\,\,\,\,\,n=9-4+1=6\)
\(\,\,\,\,\,a_1=3(4)-5=7\)
\(\,\,\,\,\,a_6=3(9)-5=22\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_6=\frac{6}{2}(7+22)\)
\(\,\,\,\,\,S_6=3(29)\)
\(\,\,\,\,\,S_6=87\)
\(\,\,\,\,\,\displaystyle\sum_{i=4}^{9}3i-5\)
\(\,\,\,\,\,n=9-4+1=6\)
\(\,\,\,\,\,a_1=3(4)-5=7\)
\(\,\,\,\,\,a_6=3(9)-5=22\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_6=\frac{6}{2}(7+22)\)
\(\,\,\,\,\,S_6=3(29)\)
\(\,\,\,\,\,S_6=87\)
\(\textbf{3)}\) \(a_1=3,\, d=5,\,\) what are \(a_8\) and \(S_8\)? The answer is \(a_8=38,\, S_8=164\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,a_8=3+(8-1)5\)
\(\,\,\,\,\,a_8=3+35\)
\(\,\,\,\,\,a_8=38\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_8=\frac{8}{2}(3+38)\)
\(\,\,\,\,\,S_8=4(41)\)
\(\,\,\,\,\,S_8=164\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,a_8=3+(8-1)5\)
\(\,\,\,\,\,a_8=3+35\)
\(\,\,\,\,\,a_8=38\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_8=\frac{8}{2}(3+38)\)
\(\,\,\,\,\,S_8=4(41)\)
\(\,\,\,\,\,S_8=164\)
\(\textbf{4)}\) \(a_1=4,\, a_5=10,\,\) what are \(d\) and \(S_8\)? The answer is \(d=1.5,\, S_8=74\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,10=4+(5-1)d\)
\(\,\,\,\,\,10=4+4d\)
\(\,\,\,\,\,6=4d\)
\(\,\,\,\,\,d=1.5\)
\(\,\,\,\,\,a_8=4+(8-1)(1.5)\)
\(\,\,\,\,\,a_8=4+10.5=14.5\)
\(\,\,\,\,\,S_8=\frac{8}{2}(4+14.5)\)
\(\,\,\,\,\,S_8=4(18.5)\)
\(\,\,\,\,\,S_8=74\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,10=4+(5-1)d\)
\(\,\,\,\,\,10=4+4d\)
\(\,\,\,\,\,6=4d\)
\(\,\,\,\,\,d=1.5\)
\(\,\,\,\,\,a_8=4+(8-1)(1.5)\)
\(\,\,\,\,\,a_8=4+10.5=14.5\)
\(\,\,\,\,\,S_8=\frac{8}{2}(4+14.5)\)
\(\,\,\,\,\,S_8=4(18.5)\)
\(\,\,\,\,\,S_8=74\)
\(\textbf{5)}\) \(a_6=22,\, S_6=90,\,\) what is \(a_1\)? The answer is \(a_1=8\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,90=\frac{6}{2}(a_1+22)\)
\(\,\,\,\,\,90=3(a_1+22)\)
\(\,\,\,\,\,30=a_1+22\)
\(\,\,\,\,\,8=a_1\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,90=\frac{6}{2}(a_1+22)\)
\(\,\,\,\,\,90=3(a_1+22)\)
\(\,\,\,\,\,30=a_1+22\)
\(\,\,\,\,\,8=a_1\)
\(\textbf{6)}\) \(3+9+15+ \cdots +75=\) The answer is \(507\)
\(\,\,\,\,\,a_1=3\)
\(\,\,\,\,\,d=9-3=6\)
\(\,\,\,\,\,a_n=75\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,75=3+(n-1)6\)
\(\,\,\,\,\,72=6(n-1)\)
\(\,\,\,\,\,12=n-1\)
\(\,\,\,\,\,13=n\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{13}=\frac{13}{2}(3+75)\)
\(\,\,\,\,\,S_{13}=\frac{13}{2}(78)\)
\(\,\,\,\,\,S_{13}=13(39)\)
\(\,\,\,\,\,S_{13}=507\)
\(\,\,\,\,\,a_1=3\)
\(\,\,\,\,\,d=9-3=6\)
\(\,\,\,\,\,a_n=75\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,75=3+(n-1)6\)
\(\,\,\,\,\,72=6(n-1)\)
\(\,\,\,\,\,12=n-1\)
\(\,\,\,\,\,13=n\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{13}=\frac{13}{2}(3+75)\)
\(\,\,\,\,\,S_{13}=\frac{13}{2}(78)\)
\(\,\,\,\,\,S_{13}=13(39)\)
\(\,\,\,\,\,S_{13}=507\)
\(\textbf{7)}\) Find the sum of the first 18 terms of \(5,8,11,14,17…\)
The Answer is \( 549\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{18}=\frac{18}{2}\left(2(5)+(18-1)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(10+(17)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(10+51\right)\)
\(\,\,\,\,\,S_{18}=9\left(61\right)\)
\(\,\,\,\,\,S_{18}=549\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{18}=\frac{18}{2}\left(2(5)+(18-1)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(10+(17)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(10+51\right)\)
\(\,\,\,\,\,S_{18}=9\left(61\right)\)
\(\,\,\,\,\,S_{18}=549\)
\(\textbf{8)}\) Find the sum of the first 12 terms of \(-6,-4,-2,0,2…\)
The Answer is \( 60\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{12}=\frac{12}{2}\left(2(-6)+(12-1)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(-12+(11)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(-12+22\right)\)
\(\,\,\,\,\,S_{12}=6\left(10\right)\)
\(\,\,\,\,\,S_{12}=60\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{12}=\frac{12}{2}\left(2(-6)+(12-1)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(-12+(11)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(-12+22\right)\)
\(\,\,\,\,\,S_{12}=6\left(10\right)\)
\(\,\,\,\,\,S_{12}=60\)
\(\textbf{9)}\) Find the sum of the first 30 terms of \(-10,-7,-4,-1,2…\)
The Answer is \( 1005\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{30}=\frac{30}{2}\left(2(-10)+(30-1)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(-20+(29)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(-20+87\right)\)
\(\,\,\,\,\,S_{30}=15\left(67\right)\)
\(\,\,\,\,\,S_{30}=1005\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{30}=\frac{30}{2}\left(2(-10)+(30-1)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(-20+(29)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(-20+87\right)\)
\(\,\,\,\,\,S_{30}=15\left(67\right)\)
\(\,\,\,\,\,S_{30}=1005\)
\(\textbf{10)}\) The sum of the first two terms is \(8\), the sum of the first three terms is \(15\), what is the value of the first term, \(a_1\)? The answer is \(a_1= 3\)
\(\text{Simplify }S_2 \text{ and }S_3\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_2=8=\frac{2}{2}\left(2a_1+(2-1)d\right)\)
\(\,\,\,\,\,S_2=8=2a_1+d\)
\(\,\,\,\,\,S_3=15=\frac{3}{2}\left(2a_1+(3-1)d\right)\)
\(\,\,\,\,\,S_3=15=\frac{3}{2}\left(2a_1+2d\right)\)
\(\,\,\,\,\,S_3=15=3a_1+3d\)
\(\text{Subtract } 3 \cdot S_2-S_3\)
\(\,\,\,\,\,3(8)-(15)=3 \cdot \left(2a_1+d\right)-\left(3a_1+3d\right)\)
\(\,\,\,\,\,24-15=6a_1+3d-3a_1-3d\)
\(\,\,\,\,\,9=3a_1\)
\(\,\,\,\,\,3=a_1\)
\(\text{Simplify }S_2 \text{ and }S_3\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_2=8=\frac{2}{2}\left(2a_1+(2-1)d\right)\)
\(\,\,\,\,\,S_2=8=2a_1+d\)
\(\,\,\,\,\,S_3=15=\frac{3}{2}\left(2a_1+(3-1)d\right)\)
\(\,\,\,\,\,S_3=15=\frac{3}{2}\left(2a_1+2d\right)\)
\(\,\,\,\,\,S_3=15=3a_1+3d\)
\(\text{Subtract } 3 \cdot S_2-S_3\)
\(\,\,\,\,\,3(8)-(15)=3 \cdot \left(2a_1+d\right)-\left(3a_1+3d\right)\)
\(\,\,\,\,\,24-15=6a_1+3d-3a_1-3d\)
\(\,\,\,\,\,9=3a_1\)
\(\,\,\,\,\,3=a_1\)
\(\textbf{11)}\) \(\displaystyle\sum_{i=2}^{6}5+4i\)
The answer is \(105\)
\(\text{Expand the summation:}\)
\(\,\,\,\,\,(5+4(2)) + (5+4(3)) + (5+4(4)) + (5+4(5)) + (5+4(6))\)
\(\,\,\,\,\,= (13) + (17) + (21) + (25) + (29)\)
\(\,\,\,\,\,= 105\)
\(\text{Expand the summation:}\)
\(\,\,\,\,\,(5+4(2)) + (5+4(3)) + (5+4(4)) + (5+4(5)) + (5+4(6))\)
\(\,\,\,\,\,= (13) + (17) + (21) + (25) + (29)\)
\(\,\,\,\,\,= 105\)
\(\textbf{12)}\) \(\displaystyle\sum_{n=1}^{7}2n-3\)
The answer is \(35\)
\(\text{Expand the summation:}\)
\(\,\,\,\,\,(2(1)-3) + (2(2)-3) + (2(3)-3) + \cdots + (2(7)-3)\)
\(\,\,\,\,\,= (-1) + (1) + (3) + (5) + (7) + (9) + (11)\)
\(\,\,\,\,\,= 35\)
\(\text{Expand the summation:}\)
\(\,\,\,\,\,(2(1)-3) + (2(2)-3) + (2(3)-3) + \cdots + (2(7)-3)\)
\(\,\,\,\,\,= (-1) + (1) + (3) + (5) + (7) + (9) + (11)\)
\(\,\,\,\,\,= 35\)
\(\textbf{13)}\) \(a_1=2,\, d=4,\,\) find \(a_{10}\) and \(S_{10}\).
The answer is \(a_{10}=38,\, S_{10}=200\)
\(\text{Find } a_{10}:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,a_{10} = 2 + (10-1)4\)
\(\,\,\,\,\,a_{10} = 2 + 36 = 38\)
\(\text{Find } S_{10}:\)
\(\,\,\,\,\,S_n = \frac{n}{2} (2a_1 + (n-1)d)\)
\(\,\,\,\,\,S_{10} = \frac{10}{2} (2(2) + (10-1)4)\)
\(\,\,\,\,\,S_{10} = 5 (4 + 36)\)
\(\,\,\,\,\,S_{10} = 5 (40) = 200\)
\(\text{Find } a_{10}:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,a_{10} = 2 + (10-1)4\)
\(\,\,\,\,\,a_{10} = 2 + 36 = 38\)
\(\text{Find } S_{10}:\)
\(\,\,\,\,\,S_n = \frac{n}{2} (2a_1 + (n-1)d)\)
\(\,\,\,\,\,S_{10} = \frac{10}{2} (2(2) + (10-1)4)\)
\(\,\,\,\,\,S_{10} = 5 (4 + 36)\)
\(\,\,\,\,\,S_{10} = 5 (40) = 200\)
\(\textbf{14)}\) \(a_1=7,\, a_6=22,\,\) find \(d\) and \(S_6\).
The answer is \(d=3,\, S_6=87\)
\(\text{Find } d:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,22 = 7 + (6-1)d\)
\(\,\,\,\,\,22 = 7 + 5d\)
\(\,\,\,\,\,5d = 15\)
\(\,\,\,\,\,d = 3\)
\(\text{Find } S_6:\)
\(\,\,\,\,\,S_n = \frac{n}{2} (2a_1 + (n-1)d)\)
\(\,\,\,\,\,S_6 = \frac{6}{2} (2(7) + (6-1)3)\)
\(\,\,\,\,\,S_6 = 3 (14 + 15)\)
\(\,\,\,\,\,S_6 = 3 (29) = 87\)
\(\text{Find } d:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,22 = 7 + (6-1)d\)
\(\,\,\,\,\,22 = 7 + 5d\)
\(\,\,\,\,\,5d = 15\)
\(\,\,\,\,\,d = 3\)
\(\text{Find } S_6:\)
\(\,\,\,\,\,S_n = \frac{n}{2} (2a_1 + (n-1)d)\)
\(\,\,\,\,\,S_6 = \frac{6}{2} (2(7) + (6-1)3)\)
\(\,\,\,\,\,S_6 = 3 (14 + 15)\)
\(\,\,\,\,\,S_6 = 3 (29) = 87\)
\(\textbf{15)}\) \(a_8=50,\, S_8=200,\,\) find \(a_1\).
The answer is \(a_1=0\)
\(\text{Use the sum formula:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,200 = \frac{8}{2} (a_1 + 50)\)
\(\,\,\,\,\,200 = 4(a_1 + 50)\)
\(\,\,\,\,\,50 = a_1 + 50\)
\(\,\,\,\,\,a_1 = 0\)
\(\text{Use the sum formula:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,200 = \frac{8}{2} (a_1 + 50)\)
\(\,\,\,\,\,200 = 4(a_1 + 50)\)
\(\,\,\,\,\,50 = a_1 + 50\)
\(\,\,\,\,\,a_1 = 0\)
\(\textbf{16)}\) \(5+10+15+\cdots+50=\)
The answer is \(275\)
\(\text{Find } n:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,50 = 5 + (n-1)5\)
\(\,\,\,\,\,45 = (n-1)5\)
\(\,\,\,\,\,9=n-1\)
\(\,\,\,\,\,10=n\)
\(\text{Find the sum:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,S_{10} = \frac{10}{2} (5 + 50)\)
\(\,\,\,\,\,S_{10} = 5 (55)\)
\(\,\,\,\,\,S_{10} = 275\)
\(\text{Find } n:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,50 = 5 + (n-1)5\)
\(\,\,\,\,\,45 = (n-1)5\)
\(\,\,\,\,\,9=n-1\)
\(\,\,\,\,\,10=n\)
\(\text{Find the sum:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,S_{10} = \frac{10}{2} (5 + 50)\)
\(\,\,\,\,\,S_{10} = 5 (55)\)
\(\,\,\,\,\,S_{10} = 275\)
\(\textbf{17)}\) Find the sum of the first 20 terms of \(4,10,16,22,\cdots\).
The answer is \(1220\)
\(\,\,\,\,\,a_1=4\)
\(\,\,\,\,\,d=10-4=6\)
\(\,\,\,\,\,n=20\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{20}=\frac{20}{2}\left(2(4)+(20-1)6\right)\)
\(\,\,\,\,\,S_{20}=10\left(8+114\right)\)
\(\,\,\,\,\,S_{20}=10(122)\)
\(\,\,\,\,\,S_{20}=1220\)
\(\,\,\,\,\,a_1=4\)
\(\,\,\,\,\,d=10-4=6\)
\(\,\,\,\,\,n=20\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{20}=\frac{20}{2}\left(2(4)+(20-1)6\right)\)
\(\,\,\,\,\,S_{20}=10\left(8+114\right)\)
\(\,\,\,\,\,S_{20}=10(122)\)
\(\,\,\,\,\,S_{20}=1220\)
\(\textbf{18)}\) \(\displaystyle\sum_{k=1}^{10}(7k+2)\)
The answer is \(405\)
\(\,\,\,\,\,a_1=7(1)+2=9\)
\(\,\,\,\,\,a_{10}=7(10)+2=72\)
\(\,\,\,\,\,n=10\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{10}=\frac{10}{2}(9+72)\)
\(\,\,\,\,\,S_{10}=5(81)\)
\(\,\,\,\,\,S_{10}=405\)
\(\,\,\,\,\,a_1=7(1)+2=9\)
\(\,\,\,\,\,a_{10}=7(10)+2=72\)
\(\,\,\,\,\,n=10\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{10}=\frac{10}{2}(9+72)\)
\(\,\,\,\,\,S_{10}=5(81)\)
\(\,\,\,\,\,S_{10}=405\)
\(\textbf{19)}\) \(a_1=-12,\, d=4,\,\) find \(a_{15}\) and \(S_{15}\).
The answer is \(a_{15}=44,\, S_{15}=240\)
\(\text{Find }a_{15}:\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,a_{15}=-12+(15-1)4\)
\(\,\,\,\,\,a_{15}=-12+56\)
\(\,\,\,\,\,a_{15}=44\)
\(\text{Find }S_{15}:\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{15}=\frac{15}{2}(-12+44)\)
\(\,\,\,\,\,S_{15}=\frac{15}{2}(32)\)
\(\,\,\,\,\,S_{15}=15(16)\)
\(\,\,\,\,\,S_{15}=240\)
\(\text{Find }a_{15}:\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,a_{15}=-12+(15-1)4\)
\(\,\,\,\,\,a_{15}=-12+56\)
\(\,\,\,\,\,a_{15}=44\)
\(\text{Find }S_{15}:\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{15}=\frac{15}{2}(-12+44)\)
\(\,\,\,\,\,S_{15}=\frac{15}{2}(32)\)
\(\,\,\,\,\,S_{15}=15(16)\)
\(\,\,\,\,\,S_{15}=240\)
See Related Pages
\(\bullet\text{ Arithmetic Sequences}\)
\(\,\,\,\,\,\,\,a_n=a_1 + d(n-1)\)
\(\bullet\text{ Geometric Sequences}\)
\(\,\,\,\,\,\,\,a_n=a_1 \cdot r^{(n-1)}…\)
\(\bullet\text{ Geometric Series}\)
\(\,\,\,\,\,\,\,s_n=a_1 \frac{1-r^n}{1-r}…\)
\(\bullet\text{ Infinite Geometric Series}\)
\(\,\,\,\,\,\,\,s_\infty = \frac{a_1}{1-r}\,\,\, |r| \lt 1…\)
\(\bullet\text{ Summation Notation}\)
\(\,\,\,\,\,\,\, \displaystyle \sum_{i=4}^{9} 3i-5 …\)
\(\bullet\text{ Recursive Sequences}\)
\(\,\,\,\,\,\,\, a_{1}=2, \, a_{n+1}=a_{n}+3…\)