Sure, let's find the sums of the given arithmetic sequences step-by-step using the approach in Gauss's Problem.
### Part a: [tex]\(1 + 2 + 3 + 4 + \ldots + 998\)[/tex]
The sum of the first [tex]\(n\)[/tex] natural numbers can be calculated using the formula:
[tex]\[
S = \frac{n(n+1)}{2}
\][/tex]
Here, [tex]\(n = 998\)[/tex]. Plugging in the value, we get:
[tex]\[
S = \frac{998 \cdot (998 + 1)}{2} = \frac{998 \cdot 999}{2} = 498501
\][/tex]
So, the sum of the sequence [tex]\(1 + 2 + 3 + \ldots + 998\)[/tex] is [tex]\(498501\)[/tex].
### Part b: [tex]\(1 + 3 + 5 + 7 + \ldots + 1001\)[/tex]
This is an arithmetic sequence where the first term [tex]\(a = 1\)[/tex], the common difference [tex]\(d = 2\)[/tex], and the last term [tex]\(l = 1001\)[/tex].
First, we find the number of terms [tex]\(n\)[/tex] in the sequence using:
[tex]\[
l = a + (n-1)d \Rightarrow 1001 = 1 + (n-1)\cdot2 \Rightarrow 1000 = 2(n-1) \Rightarrow n = 501
\][/tex]
The sum of an arithmetic sequence is given by:
[tex]\[
S = \frac{n(a + l)}{2}
\][/tex]
Substituting the values, we get:
[tex]\[
S = \frac{501(1 + 1001)}{2} = \frac{501 \cdot 1002}{2} = 251001
\][/tex]
So, the sum of the sequence [tex]\(1 + 3 + 5 + \ldots + 1001\)[/tex] is [tex]\(251001\)[/tex].
### Part c: [tex]\(2 + 5 + 8 + 11 + \ldots + 293\)[/tex]
This is an arithmetic sequence where the first term [tex]\(a = 2\)[/tex], the common difference [tex]\(d = 3\)[/tex], and the last term [tex]\(l = 293\)[/tex].
First, we find the number of terms [tex]\(n\)[/tex] in the sequence using:
[tex]\[
l = a + (n-1)d \Rightarrow 293 = 2 + (n-1)\cdot3 \Rightarrow 291 = 3(n-1) \Rightarrow n = 98
\][/tex]
The sum of an arithmetic sequence is:
[tex]\[
S = \frac{n(a + l)}{2}
\][/tex]
Substituting the values, we get:
[tex]\[
S = \frac{98(2 + 293)}{2} = \frac{98 \cdot 295}{2} = 14455
\][/tex]
So, the sum of the sequence [tex]\(2 + 5 + 8 + \ldots + 293\)[/tex] is [tex]\(14455\)[/tex].
### Part d: [tex]\(685 + 678 + 671 + 664 + \ldots + 6\)[/tex]
This is an arithmetic sequence where the first term [tex]\(a = 685\)[/tex], the common difference [tex]\(d = -7\)[/tex], and the last term [tex]\(l = 6\)[/tex].
First, we find the number of terms [tex]\(n\)[/tex] in the sequence using:
[tex]\[
l = a + (n-1)d \Rightarrow 6 = 685 + (n-1)(-7) \Rightarrow 6 = 685 - 7(n-1) \Rightarrow -679 = -7(n-1) \Rightarrow 679 = 7(n-1) \Rightarrow n = 98
\][/tex]
The sum of an arithmetic sequence is:
[tex]\[
S = \frac{n(a + l)}{2}
\][/tex]
Substituting the values, we get:
[tex]\[
S = \frac{98(685 + 6)}{2} = \frac{98 \cdot 691}{2} = 33859
\][/tex]
So, the sum of the sequence [tex]\(685 + 678 + 671 + \ldots + 6\)[/tex] is [tex]\(33859\)[/tex].
Overall, the sums are:
- [tex]\(1 + 2 + 3 + \ldots + 998 = 498501\)[/tex]
- [tex]\(1 + 3 + 5 + \ldots + 1001 = 251001\)[/tex]
- [tex]\(2 + 5 + 8 + \ldots + 293 = 14455\)[/tex]
- [tex]\(685 + 678 + 671 + \ldots + 6 = 33859\)[/tex]
