Connect with a community that values knowledge and expertise on IDNLearn.com. Join our interactive community and get comprehensive, reliable answers to all your questions.
Sagot :
Let's first define the general terms for both the arithmetic sequence and the geometric sequence based on the information provided in the problem.
### Arithmetic Sequence
The first term, [tex]\(a_1\)[/tex], is 400, and the common difference, [tex]\(d\)[/tex], is 500. The [tex]\(n\)[/tex]th term of an arithmetic sequence can be expressed as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
Substituting the given values:
[tex]\[ a_n = 400 + (n-1) \cdot 500 \][/tex]
[tex]\[ a_n = 400 + 500n - 500 \][/tex]
[tex]\[ a_n = 500n - 100 \][/tex]
### Geometric Sequence
The first term, [tex]\(g_1\)[/tex], is 3, and the common ratio, [tex]\(r\)[/tex], is 3. The [tex]\(n\)[/tex]th term of a geometric sequence can be expressed as:
[tex]\[ g_n = g_1 \cdot r^{n-1} \][/tex]
Substituting the given values:
[tex]\[ g_n = 3 \cdot 3^{n-1} \][/tex]
[tex]\[ g_n = 3^n \][/tex]
### Finding [tex]\(n\)[/tex] such that [tex]\( g_n > a_n \)[/tex]
We need to find the smallest [tex]\(n\)[/tex] for which the [tex]\(n\)[/tex]th term of the geometric sequence exceeds the [tex]\(n\)[/tex]th term of the arithmetic sequence, i.e., [tex]\( g_n > a_n \)[/tex].
Set the expressions for [tex]\(a_n\)[/tex] and [tex]\(g_n\)[/tex] and solve for [tex]\(n\)[/tex]:
[tex]\[ 3^n > 500n - 100 \][/tex]
### Evaluating terms
Let's evaluate the sequences term by term until we find when the geometric sequence term first exceeds the arithmetic sequence term:
1. For [tex]\(n = 1\)[/tex]:
[tex]\[ a_1 = 500 \cdot 1 - 100 = 400 \][/tex]
[tex]\[ g_1 = 3^1 = 3 \][/tex]
Clearly, [tex]\(3 < 400\)[/tex].
2. For [tex]\(n = 2\)[/tex]:
[tex]\[ a_2 = 500 \cdot 2 - 100 = 900 \][/tex]
[tex]\[ g_2 = 3^2 = 9 \][/tex]
Clearly, [tex]\(9 < 900\)[/tex].
3. For [tex]\(n = 3\)[/tex]:
[tex]\[ a_3 = 500 \cdot 3 - 100 = 1400 \][/tex]
[tex]\[ g_3 = 3^3 = 27 \][/tex]
Clearly, [tex]\(27 < 1400\)[/tex].
4. For [tex]\(n = 4\)[/tex]:
[tex]\[ a_4 = 500 \cdot 4 - 100 = 1900 \][/tex]
[tex]\[ g_4 = 3^4 = 81 \][/tex]
Clearly, [tex]\(81 < 1900\)[/tex].
5. For [tex]\(n = 5\)[/tex]:
[tex]\[ a_5 = 500 \cdot 5 - 100 = 2400 \][/tex]
[tex]\[ g_5 = 3^5 = 243 \][/tex]
Clearly, [tex]\(243 < 2400\)[/tex].
6. For [tex]\(n = 6\)[/tex]:
[tex]\[ a_6 = 500 \cdot 6 - 100 = 2900 \][/tex]
[tex]\[ g_6 = 3^6 = 729 \][/tex]
Clearly, [tex]\(729 < 2900\)[/tex].
7. For [tex]\(n = 7\)[/tex]:
[tex]\[ a_7 = 500 \cdot 7 - 100 = 3400 \][/tex]
[tex]\[ g_7 = 3^7 = 2187 \][/tex]
Clearly, [tex]\(2187 < 3400\)[/tex].
8. For [tex]\(n = 8\)[/tex]:
[tex]\[ a_8 = 500 \cdot 8 - 100 = 3900 \][/tex]
[tex]\[ g_8 = 3^8 = 6561 \][/tex]
Here, [tex]\(6561 > 3900\)[/tex].
So, the least number [tex]\(n\)[/tex] such that the [tex]\(n\)[/tex]th term of the geometric sequence is greater than the corresponding term in the arithmetic sequence is:
[tex]\[ n = 8 \][/tex]
Therefore, the geometric sequence is larger than the arithmetic sequence at the [tex]\(8\)[/tex]th term.
### Arithmetic Sequence
The first term, [tex]\(a_1\)[/tex], is 400, and the common difference, [tex]\(d\)[/tex], is 500. The [tex]\(n\)[/tex]th term of an arithmetic sequence can be expressed as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
Substituting the given values:
[tex]\[ a_n = 400 + (n-1) \cdot 500 \][/tex]
[tex]\[ a_n = 400 + 500n - 500 \][/tex]
[tex]\[ a_n = 500n - 100 \][/tex]
### Geometric Sequence
The first term, [tex]\(g_1\)[/tex], is 3, and the common ratio, [tex]\(r\)[/tex], is 3. The [tex]\(n\)[/tex]th term of a geometric sequence can be expressed as:
[tex]\[ g_n = g_1 \cdot r^{n-1} \][/tex]
Substituting the given values:
[tex]\[ g_n = 3 \cdot 3^{n-1} \][/tex]
[tex]\[ g_n = 3^n \][/tex]
### Finding [tex]\(n\)[/tex] such that [tex]\( g_n > a_n \)[/tex]
We need to find the smallest [tex]\(n\)[/tex] for which the [tex]\(n\)[/tex]th term of the geometric sequence exceeds the [tex]\(n\)[/tex]th term of the arithmetic sequence, i.e., [tex]\( g_n > a_n \)[/tex].
Set the expressions for [tex]\(a_n\)[/tex] and [tex]\(g_n\)[/tex] and solve for [tex]\(n\)[/tex]:
[tex]\[ 3^n > 500n - 100 \][/tex]
### Evaluating terms
Let's evaluate the sequences term by term until we find when the geometric sequence term first exceeds the arithmetic sequence term:
1. For [tex]\(n = 1\)[/tex]:
[tex]\[ a_1 = 500 \cdot 1 - 100 = 400 \][/tex]
[tex]\[ g_1 = 3^1 = 3 \][/tex]
Clearly, [tex]\(3 < 400\)[/tex].
2. For [tex]\(n = 2\)[/tex]:
[tex]\[ a_2 = 500 \cdot 2 - 100 = 900 \][/tex]
[tex]\[ g_2 = 3^2 = 9 \][/tex]
Clearly, [tex]\(9 < 900\)[/tex].
3. For [tex]\(n = 3\)[/tex]:
[tex]\[ a_3 = 500 \cdot 3 - 100 = 1400 \][/tex]
[tex]\[ g_3 = 3^3 = 27 \][/tex]
Clearly, [tex]\(27 < 1400\)[/tex].
4. For [tex]\(n = 4\)[/tex]:
[tex]\[ a_4 = 500 \cdot 4 - 100 = 1900 \][/tex]
[tex]\[ g_4 = 3^4 = 81 \][/tex]
Clearly, [tex]\(81 < 1900\)[/tex].
5. For [tex]\(n = 5\)[/tex]:
[tex]\[ a_5 = 500 \cdot 5 - 100 = 2400 \][/tex]
[tex]\[ g_5 = 3^5 = 243 \][/tex]
Clearly, [tex]\(243 < 2400\)[/tex].
6. For [tex]\(n = 6\)[/tex]:
[tex]\[ a_6 = 500 \cdot 6 - 100 = 2900 \][/tex]
[tex]\[ g_6 = 3^6 = 729 \][/tex]
Clearly, [tex]\(729 < 2900\)[/tex].
7. For [tex]\(n = 7\)[/tex]:
[tex]\[ a_7 = 500 \cdot 7 - 100 = 3400 \][/tex]
[tex]\[ g_7 = 3^7 = 2187 \][/tex]
Clearly, [tex]\(2187 < 3400\)[/tex].
8. For [tex]\(n = 8\)[/tex]:
[tex]\[ a_8 = 500 \cdot 8 - 100 = 3900 \][/tex]
[tex]\[ g_8 = 3^8 = 6561 \][/tex]
Here, [tex]\(6561 > 3900\)[/tex].
So, the least number [tex]\(n\)[/tex] such that the [tex]\(n\)[/tex]th term of the geometric sequence is greater than the corresponding term in the arithmetic sequence is:
[tex]\[ n = 8 \][/tex]
Therefore, the geometric sequence is larger than the arithmetic sequence at the [tex]\(8\)[/tex]th term.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.
