+71 1. Write a rule for the nth term of the arithmetic sequence with a1= -7 and the common difference of 5/2.
2. Find the sum of the first 10 terms of the arithmetic series. -5 + 0 + 5 + 10 +...
a.350 b.170 c.175 d.180
3. Find the sum of the first 22 terms of the arithmetic sequence, if the first term is –2 and the common difference is –5.
a.–1199 b.545 c.–1177 d.–2398
4. Find the common difference of the arithmetic sequence.
0, 0.4, 0.8, 1.2, . . .
a.–0.3 b.0.3 c.0.4 d. –0.4
5. Tell whether the sequence is arithmetic. If it finds the common difference.
a. -5, -1, 3, 7, 11,...
b. -1, -1/3, 1/3, 1 , 5/3,...
6. Identify the sequence as arithmetic, geometric, or neither.
1, 1, 2, 3, 5, 8, 13, . . .
+2440 Arithmetic sequences are written in the form of the following:
\(a_n=a_1+d(n-1)\).
an is the nth term of the sequence
a1 is the first term of the sequence
d is the common difference
n is the desired term number
1) We already know the information necessary to write an equation for the nth term of this sequence.
| \(a_1=-7; d=\frac{5}{2}\\ a_n=-7+\frac{5}{2}(n-1)\) | The only thing left to do is simplify. Distributing is the first step to accomplish this. |
| \(a_n=-\frac{14}{2}+\frac{5}{2}n-\frac{5}{2}\) | Combine like terms. |
| \(a_n=\frac{5}{2}n-\frac{19}{2}\) | This is completely simplified. |
2) There is a formula that exists for the summation of an arithmetic series or geometric series, but that probably would not help your understanding anyway; I can derive it for you, though.
We have to know the last term, and we can generate this by using the formula from before: \(a_n=a_1+d(n-1)\). Therefore, n=10, d=5, and a1=5:
| \(a_{10}=-5+5(10-1)\) | Evaluate this to determine the last term of the sequence. |
| \(a_{10}=-5+5*9\) | Simplify the right hand side. |
| \(a_{10}=40\) | |
Now, let's attempt to evaluate the sum. Standard notation dictates \(S_n\) for the summation.
| \(S_{10}=-5+0+...+35+40\\ S_{10}=\hspace{2mm}40+35+...+0\hspace{2mm}-5\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ 2S_{10}=\underbrace{35+35+...+35+35}=35*10=350\\ \hspace{24mm}\text{10 times}\\ S_{10}=175\) | All I did here is reverse the same sum and added both of them together. This made it significantly easier to determine the sum. |
I did not answer every question here because the others can be answered with the knowledge given above, albeit not directly. Try and figure it out yourself.
