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 \(-15\)
\(\,\,\,\,\,\displaystyle\sum_{i=3}^{5}3-2i\)
\(\,\,\,\,\,=(3-2(3))+(3-2(4))+(3-2(5))\)
\(\,\,\,\,\,=(3-6)+(3-8)+(3-10)\)
\(\,\,\,\,\,=-3+(-5)+(-7)\)
\(\,\,\,\,\,=-15\)
\(\textbf{2)}\) \(\displaystyle\sum_{i=4}^{9}3i-5\)
The answer is \(87\)
\(\,\,\,\,\,\displaystyle\sum_{i=4}^{9}3i-5\)
\(\,\,\,\,\,n=9-4+1=6\)
\(\,\,\,\,\,a_1=3(4)-5=7\)
\(\,\,\,\,\,a_6=3(9)-5=22\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_6=\frac{6}{2}(7+22)\)
\(\,\,\,\,\,S_6=3(29)\)
\(\,\,\,\,\,S_6=87\)
\(\textbf{3)}\) \(a_1=3,\, d=5,\,\) what are \(a_8\) and \(S_8\)?
The answer is \(a_8=38,\, S_8=164\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,a_8=3+(8-1)5\)
\(\,\,\,\,\,a_8=3+35\)
\(\,\,\,\,\,a_8=38\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_8=\frac{8}{2}(3+38)\)
\(\,\,\,\,\,S_8=4(41)\)
\(\,\,\,\,\,S_8=164\)
\(\textbf{4)}\) \(a_1=4,\, a_5=10,\,\) what are \(d\) and \(S_8\)?
The answer is \(d=1.5,\, S_8=74\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,10=4+(5-1)d\)
\(\,\,\,\,\,10=4+4d\)
\(\,\,\,\,\,6=4d\)
\(\,\,\,\,\,d=1.5\)
\(\,\,\,\,\,a_8=4+(8-1)(1.5)\)
\(\,\,\,\,\,a_8=4+10.5=14.5\)
\(\,\,\,\,\,S_8=\frac{8}{2}(4+14.5)\)
\(\,\,\,\,\,S_8=4(18.5)\)
\(\,\,\,\,\,S_8=74\)
\(\textbf{5)}\) \(a_6=22,\, S_6=90,\,\) what is \(a_1\)?
The answer is \(a_1=8\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,90=\frac{6}{2}(a_1+22)\)
\(\,\,\,\,\,90=3(a_1+22)\)
\(\,\,\,\,\,30=a_1+22\)
\(\,\,\,\,\,8=a_1\)
\(\textbf{6)}\) \(3+9+15+ \cdots +75=\)
The answer is \(507\)
\(\,\,\,\,\,a_1=3\)
\(\,\,\,\,\,d=9-3=6\)
\(\,\,\,\,\,a_n=75\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,75=3+(n-1)6\)
\(\,\,\,\,\,72=6(n-1)\)
\(\,\,\,\,\,12=n-1\)
\(\,\,\,\,\,13=n\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{13}=\frac{13}{2}(3+75)\)
\(\,\,\,\,\,S_{13}=\frac{13}{2}(78)\)
\(\,\,\,\,\,S_{13}=13(39)\)
\(\,\,\,\,\,S_{13}=507\)
\(\textbf{7)}\) Find the sum of the first 18 terms of \(5,8,11,14,17…\)
The Answer is \( 549\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{18}=\frac{18}{2}\left(2(5)+(18-1)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(10+(17)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(10+51\right)\)
\(\,\,\,\,\,S_{18}=9\left(61\right)\)
\(\,\,\,\,\,S_{18}=549\)
\(\textbf{8)}\) Find the sum of the first 12 terms of \(-6,-4,-2,0,2…\)
The Answer is \( 60\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{12}=\frac{12}{2}\left(2(-6)+(12-1)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(-12+(11)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(-12+22\right)\)
\(\,\,\,\,\,S_{12}=6\left(10\right)\)
\(\,\,\,\,\,S_{12}=60\)
\(\textbf{9)}\) Find the sum of the first 30 terms of \(-10,-7,-4,-1,2…\)
The Answer is \( 1005\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{30}=\frac{30}{2}\left(2(-10)+(30-1)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(-20+(29)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(-20+87\right)\)
\(\,\,\,\,\,S_{30}=15\left(67\right)\)
\(\,\,\,\,\,S_{30}=1005\)
\(\textbf{10)}\) 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 \(a_1= 3\)
\(\text{Simplify }S_2 \text{ and }S_3\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_2=8=\frac{2}{2}\left(2a_1+(2-1)d\right)\)
\(\,\,\,\,\,S_2=8=2a_1+d\)
\(\,\,\,\,\,S_3=15=\frac{3}{2}\left(2a_1+(3-1)d\right)\)
\(\,\,\,\,\,S_3=15=\frac{3}{2}\left(2a_1+2d\right)\)
\(\,\,\,\,\,S_3=15=3a_1+3d\)
\(\text{Subtract } 3 \cdot S_2-S_3\)
\(\,\,\,\,\,3(8)-(15)=3 \cdot \left(2a_1+d\right)-\left(3a_1+3d\right)\)
\(\,\,\,\,\,24-15=6a_1+3d-3a_1-3d\)
\(\,\,\,\,\,9=3a_1\)
\(\,\,\,\,\,3=a_1\)
\(\textbf{11)}\) \(\displaystyle\sum_{i=2}^{6}5+4i\)
The answer is \(105\)
\(\text{Expand the summation:}\)
\(\,\,\,\,\,(5+4(2)) + (5+4(3)) + (5+4(4)) + (5+4(5)) + (5+4(6))\)
\(\,\,\,\,\,= (13) + (17) + (21) + (25) + (29)\)
\(\,\,\,\,\,= 105\)
\(\textbf{12)}\) \(\displaystyle\sum_{n=1}^{7}2n-3\)
The answer is \(35\)
\(\text{Expand the summation:}\)
\(\,\,\,\,\,(2(1)-3) + (2(2)-3) + (2(3)-3) + \cdots + (2(7)-3)\)
\(\,\,\,\,\,= (-1) + (1) + (3) + (5) + (7) + (9) + (11)\)
\(\,\,\,\,\,= 35\)
\(\textbf{13)}\) \(a_1=2,\, d=4,\,\) find \(a_{10}\) and \(S_{10}\).
The answer is \(a_{10}=38,\, S_{10}=200\)
\(\text{Find } a_{10}:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,a_{10} = 2 + (10-1)4\)
\(\,\,\,\,\,a_{10} = 2 + 36 = 38\)
\(\text{Find } S_{10}:\)
\(\,\,\,\,\,S_n = \frac{n}{2} (2a_1 + (n-1)d)\)
\(\,\,\,\,\,S_{10} = \frac{10}{2} (2(2) + (10-1)4)\)
\(\,\,\,\,\,S_{10} = 5 (4 + 36)\)
\(\,\,\,\,\,S_{10} = 5 (40) = 200\)
\(\textbf{14)}\) \(a_1=7,\, a_6=22,\,\) find \(d\) and \(S_6\).
The answer is \(d=3,\, S_6=87\)
\(\text{Find } d:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,22 = 7 + (6-1)d\)
\(\,\,\,\,\,22 = 7 + 5d\)
\(\,\,\,\,\,5d = 15\)
\(\,\,\,\,\,d = 3\)
\(\text{Find } S_6:\)
\(\,\,\,\,\,S_n = \frac{n}{2} (2a_1 + (n-1)d)\)
\(\,\,\,\,\,S_6 = \frac{6}{2} (2(7) + (6-1)3)\)
\(\,\,\,\,\,S_6 = 3 (14 + 15)\)
\(\,\,\,\,\,S_6 = 3 (29) = 87\)
\(\textbf{15)}\) \(a_8=50,\, S_8=200,\,\) find \(a_1\).
The answer is \(a_1=0\)
\(\text{Use the sum formula:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,200 = \frac{8}{2} (a_1 + 50)\)
\(\,\,\,\,\,200 = 4(a_1 + 50)\)
\(\,\,\,\,\,50 = a_1 + 50\)
\(\,\,\,\,\,a_1 = 0\)
\(\textbf{16)}\) \(5+10+15+\cdots+50=\)
The answer is \(275\)
\(\text{Find } n:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,50 = 5 + (n-1)5\)
\(\,\,\,\,\,45 = (n-1)5\)
\(\,\,\,\,\,9=n-1\)
\(\,\,\,\,\,10=n\)
\(\text{Find the sum:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,S_{10} = \frac{10}{2} (5 + 50)\)
\(\,\,\,\,\,S_{10} = 5 (55)\)
\(\,\,\,\,\,S_{10} = 275\)
\(\textbf{17)}\) Find the sum of the first 20 terms of \(4,10,16,22,\cdots\).
The answer is \(1220\)
\(\,\,\,\,\,a_1=4\)
\(\,\,\,\,\,d=10-4=6\)
\(\,\,\,\,\,n=20\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{20}=\frac{20}{2}\left(2(4)+(20-1)6\right)\)
\(\,\,\,\,\,S_{20}=10\left(8+114\right)\)
\(\,\,\,\,\,S_{20}=10(122)\)
\(\,\,\,\,\,S_{20}=1220\)
\(\textbf{18)}\) \(\displaystyle\sum_{k=1}^{10}(7k+2)\)
The answer is \(405\)
\(\,\,\,\,\,a_1=7(1)+2=9\)
\(\,\,\,\,\,a_{10}=7(10)+2=72\)
\(\,\,\,\,\,n=10\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{10}=\frac{10}{2}(9+72)\)
\(\,\,\,\,\,S_{10}=5(81)\)
\(\,\,\,\,\,S_{10}=405\)
\(\textbf{19)}\) \(a_1=-12,\, d=4,\,\) find \(a_{15}\) and \(S_{15}\).
The answer is \(a_{15}=44,\, S_{15}=240\)
\(\text{Find }a_{15}:\)
\(\,\,\,\,\,a_n=a_1+(n-1)d\)
\(\,\,\,\,\,a_{15}=-12+(15-1)4\)
\(\,\,\,\,\,a_{15}=-12+56\)
\(\,\,\,\,\,a_{15}=44\)
\(\text{Find }S_{15}:\)
\(\,\,\,\,\,S_n=\frac{n}{2}(a_1+a_n)\)
\(\,\,\,\,\,S_{15}=\frac{15}{2}(-12+44)\)
\(\,\,\,\,\,S_{15}=\frac{15}{2}(32)\)
\(\,\,\,\,\,S_{15}=15(16)\)
\(\,\,\,\,\,S_{15}=240\)
See Related Pages
