Get comprehensive solutions to your problems with IDNLearn.com. Our community is ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
Let's find the first three terms for each sequence defined by the given general rules.
### Sequence 1: [tex]\( a_n = n + 3 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 1 + 3 = 4 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 2 + 3 = 5 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3 + 3 = 6 \][/tex]
So, the first three terms are [tex]\( 4, 5, 6 \)[/tex].
### Sequence 2: [tex]\( a_n = 2n - 4 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 2 \cdot 1 - 4 = -2 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 2 \cdot 2 - 4 = 0 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 2 \cdot 3 - 4 = 2 \][/tex]
So, the first three terms are [tex]\( -2, 0, 2 \)[/tex].
### Sequence 3: [tex]\( a_n = 3n + 10 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 3 \cdot 1 + 10 = 13 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 3 \cdot 2 + 10 = 16 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3 \cdot 3 + 10 = 19 \][/tex]
So, the first three terms are [tex]\( 13, 16, 19 \)[/tex].
### Sequence 4: [tex]\( a_n = -n + 5 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = -1 + 5 = 4 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = -2 + 5 = 3 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = -3 + 5 = 2 \][/tex]
So, the first three terms are [tex]\( 4, 3, 2 \)[/tex].
### Sequence 5: [tex]\( a_n = -4n + 6 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = -4 \cdot 1 + 6 = 2 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = -4 \cdot 2 + 6 = -2 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = -4 \cdot 3 + 6 = -6 \][/tex]
So, the first three terms are [tex]\( 2, -2, -6 \)[/tex].
### Sequence 6: [tex]\( a_n = n^2 - 2 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 1^2 - 2 = -1 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 2^2 - 2 = 2 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3^2 - 2 = 7 \][/tex]
So, the first three terms are [tex]\( -1, 2, 7 \)[/tex].
### Sequence 7: [tex]\( a_n = 3n^2 + 2n \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 3 \cdot 1^2 + 2 \cdot 1 = 3 + 2 = 5 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 3 \cdot 2^2 + 2 \cdot 2 = 3 \cdot 4 + 4 = 12 + 4 = 16 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3 \cdot 3^2 + 2 \cdot 3 = 3 \cdot 9 + 6 = 27 + 6 = 33 \][/tex]
So, the first three terms are [tex]\( 5, 16, 33 \)[/tex].
### Sequence 8: [tex]\( a_n = \frac{n + 3}{2} \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = \frac{1 + 3}{2} = \frac{4}{2} = 2.0 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = \frac{2 + 3}{2} = \frac{5}{2} = 2.5 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = \frac{3 + 3}{2} = \frac{6}{2} = 3.0 \][/tex]
So, the first three terms are [tex]\( 2.0, 2.5, 3.0 \)[/tex].
### Sequence 9: [tex]\( a_n = 3\sqrt{n + 1} \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 3\sqrt{1 + 1} = 3\sqrt{2} \approx 4.242640687119286 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 3\sqrt{2 + 1} = 3\sqrt{3} \approx 5.196152422706632 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3\sqrt{3 + 1} = 3\sqrt{4} = 3 \cdot 2 = 6.0 \][/tex]
So, the first three terms are [tex]\( 4.242640687119286, 5.196152422706632, 6.0 \)[/tex].
### Sequence 10: [tex]\( a_n = 11(n + 2)^2 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 11(1 + 2)^2 = 11 \cdot 3^2 = 11 \cdot 9 = 99 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 11(2 + 2)^2 = 11 \cdot 4^2 = 11 \cdot 16 = 176 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 11(3 + 2)^2 = 11 \cdot 5^2 = 11 \cdot 25 = 275 \][/tex]
So, the first three terms are [tex]\( 99, 176, 275 \)[/tex].
In summary, the first three terms of each sequence are:
1. [tex]\( 4, 5, 6 \)[/tex]
2. [tex]\( -2, 0, 2 \)[/tex]
3. [tex]\( 13, 16, 19 \)[/tex]
4. [tex]\( 4, 3, 2 \)[/tex]
5. [tex]\( 2, -2, -6 \)[/tex]
6. [tex]\( -1, 2, 7 \)[/tex]
7. [tex]\( 5, 16, 33 \)[/tex]
8. [tex]\( 2.0, 2.5, 3.0 \)[/tex]
9. [tex]\( 4.242640687119286, 5.196152422706632, 6.0 \)[/tex]
10. [tex]\( 99, 176, 275 \)[/tex]
### Sequence 1: [tex]\( a_n = n + 3 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 1 + 3 = 4 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 2 + 3 = 5 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3 + 3 = 6 \][/tex]
So, the first three terms are [tex]\( 4, 5, 6 \)[/tex].
### Sequence 2: [tex]\( a_n = 2n - 4 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 2 \cdot 1 - 4 = -2 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 2 \cdot 2 - 4 = 0 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 2 \cdot 3 - 4 = 2 \][/tex]
So, the first three terms are [tex]\( -2, 0, 2 \)[/tex].
### Sequence 3: [tex]\( a_n = 3n + 10 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 3 \cdot 1 + 10 = 13 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 3 \cdot 2 + 10 = 16 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3 \cdot 3 + 10 = 19 \][/tex]
So, the first three terms are [tex]\( 13, 16, 19 \)[/tex].
### Sequence 4: [tex]\( a_n = -n + 5 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = -1 + 5 = 4 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = -2 + 5 = 3 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = -3 + 5 = 2 \][/tex]
So, the first three terms are [tex]\( 4, 3, 2 \)[/tex].
### Sequence 5: [tex]\( a_n = -4n + 6 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = -4 \cdot 1 + 6 = 2 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = -4 \cdot 2 + 6 = -2 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = -4 \cdot 3 + 6 = -6 \][/tex]
So, the first three terms are [tex]\( 2, -2, -6 \)[/tex].
### Sequence 6: [tex]\( a_n = n^2 - 2 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 1^2 - 2 = -1 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 2^2 - 2 = 2 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3^2 - 2 = 7 \][/tex]
So, the first three terms are [tex]\( -1, 2, 7 \)[/tex].
### Sequence 7: [tex]\( a_n = 3n^2 + 2n \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 3 \cdot 1^2 + 2 \cdot 1 = 3 + 2 = 5 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 3 \cdot 2^2 + 2 \cdot 2 = 3 \cdot 4 + 4 = 12 + 4 = 16 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3 \cdot 3^2 + 2 \cdot 3 = 3 \cdot 9 + 6 = 27 + 6 = 33 \][/tex]
So, the first three terms are [tex]\( 5, 16, 33 \)[/tex].
### Sequence 8: [tex]\( a_n = \frac{n + 3}{2} \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = \frac{1 + 3}{2} = \frac{4}{2} = 2.0 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = \frac{2 + 3}{2} = \frac{5}{2} = 2.5 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = \frac{3 + 3}{2} = \frac{6}{2} = 3.0 \][/tex]
So, the first three terms are [tex]\( 2.0, 2.5, 3.0 \)[/tex].
### Sequence 9: [tex]\( a_n = 3\sqrt{n + 1} \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 3\sqrt{1 + 1} = 3\sqrt{2} \approx 4.242640687119286 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 3\sqrt{2 + 1} = 3\sqrt{3} \approx 5.196152422706632 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 3\sqrt{3 + 1} = 3\sqrt{4} = 3 \cdot 2 = 6.0 \][/tex]
So, the first three terms are [tex]\( 4.242640687119286, 5.196152422706632, 6.0 \)[/tex].
### Sequence 10: [tex]\( a_n = 11(n + 2)^2 \)[/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 11(1 + 2)^2 = 11 \cdot 3^2 = 11 \cdot 9 = 99 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = 11(2 + 2)^2 = 11 \cdot 4^2 = 11 \cdot 16 = 176 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = 11(3 + 2)^2 = 11 \cdot 5^2 = 11 \cdot 25 = 275 \][/tex]
So, the first three terms are [tex]\( 99, 176, 275 \)[/tex].
In summary, the first three terms of each sequence are:
1. [tex]\( 4, 5, 6 \)[/tex]
2. [tex]\( -2, 0, 2 \)[/tex]
3. [tex]\( 13, 16, 19 \)[/tex]
4. [tex]\( 4, 3, 2 \)[/tex]
5. [tex]\( 2, -2, -6 \)[/tex]
6. [tex]\( -1, 2, 7 \)[/tex]
7. [tex]\( 5, 16, 33 \)[/tex]
8. [tex]\( 2.0, 2.5, 3.0 \)[/tex]
9. [tex]\( 4.242640687119286, 5.196152422706632, 6.0 \)[/tex]
10. [tex]\( 99, 176, 275 \)[/tex]
We are happy to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
