Notes
Questions
\(\textbf{1)}\) \(\displaystyle\sum_{i=3}^{5}3-2i\)
The answer is \(-15\)
\(\textbf{2)}\) \(\displaystyle\sum_{i=4}^{9}3i-5\)
The answer is \(87\)
\(\textbf{3)}\) \(a_1=3,\, d=5,\,\) what are \(a_8\) and \(S_8\)?
The answer is \(a_8=38,\, S_8=164\)
\(\textbf{4)}\) \(a_1=4,\, a_5=10,\,\) what are \(d\) and \(S_8\)?
The answer is \(d=1.5,\, S_8=74\)
\(\textbf{5)}\) \(a_6=22,\, S_6=90,\,\) what is \(a_1\)?
The answer is \(a_1=8\)
\(\textbf{6)}\) \(3+9+15+ \cdots +75=\)
The answer is \(507\)
\(\textbf{7)}\) Find the sum of the first 18 terms of \(5,8,11,14,17…\)
The Answer is \( 549\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{18}=\frac{18}{2}\left(2(5)+(18-1)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(10+(17)3\right)\)
\(\,\,\,\,\,S_{18}=9\left(61\right)\)
\(\,\,\,\,\,549\)
\(\textbf{8)}\) Find the sum of the first 12 terms of \(-6,-4,-2,0,2…\)
The Answer is \( 60\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{12}=\frac{12}{2}\left(2(-6)+(12-1)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(-12+(11)2\right)\)
\(\,\,\,\,\,S_{12}=6\left(10\right)\)
\(\,\,\,\,\,60\)
\(\textbf{9)}\) Find the sum of the first 30 terms of \(-10,-7,-4,-1,2…\)
The Answer is \( 1005\)
\(\,\,\,\,\,S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\)
\(\,\,\,\,\,S_{30}=\frac{30}{2}\left(2(-10)+(30-1)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(-20+(29)3\right)\)
\(\,\,\,\,\,S_{30}=15\left(67\right)\)
\(\,\,\,\,\,1005\)
\(\textbf{10)}\) 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 \(a_1=8\)
\(\text{Use sum formula:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,200 = \frac{8}{2} (a_1 + 50)\)
\(\,\,\,\,\,200 = 4(a_1 + 50)\)
\(\,\,\,\,\,50 = a_1 + 50\)
\(\,\,\,\,\,a_1 = 8\)
\(\textbf{16)}\) \(5+10+15+\cdots+50=\)
The answer is \(275\)
\(\text{Find } n:\)
\(\,\,\,\,\,a_n = a_1 + (n-1)d\)
\(\,\,\,\,\,50 = 5 + (n-1)5\)
\(\,\,\,\,\,45 = (n-1)5\)
\(\,\,\,\,\,n = 10\)
\(\text{Find sum:}\)
\(\,\,\,\,\,S_n = \frac{n}{2} (a_1 + a_n)\)
\(\,\,\,\,\,S_{10} = \frac{10}{2} (5 + 50)\)
\(\,\,\,\,\,S_{10} = 5 (55)\)
\(\,\,\,\,\,S_{10} = 275\)
See Related Pages
In Summary
An arithmetic series is a sequence of numbers in which each term is the sum of the previous term and a fixed constant. The constant is called the common difference, and the sequence is called an arithmetic sequence. This is typically studied in math classes such as algebra and pre-calculus. They are an important topic in these courses because they provide a foundation for more advanced mathematical concepts, and have many real world applications.
