Question
Consider the following two sequences of years: ● Sequence 1: 1998, 2005, 2012, . . . ● Sequence 2: 1996, 2005, 2014, . . . What is the next year after 2005 that appears in both sequences
Answers
GPT-4o mini
To find the next year after 2005 that appears in both sequences, we first need to determine the general formulas for each sequence:
**Sequence 1:**
The first term is 1998 and the sequence appears to increase by 7 years each time. The terms can be expressed as:
\[
a_n = 1998 + 7(n - 1) = 1991 + 7n
\]
for \( n = 1, 2, 3, \ldots \).
Thus, the years in Sequence 1 are:
- \( a_1 = 1998 \)
- \( a_2 = 2005 \)
- \( a_3 = 2012 \)
- \( a_4 = 2019 \)
- \( a_5 = 2026 \)
- \( a_6 = 2033 \), and so on.
**Sequence 2:**
The first term is 1996 and the sequence also appears to increase by 9 years each time. The terms can be expressed as:
\[
b_n = 1996 + 9(n - 1) = 1987 + 9n
\]
for \( n = 1, 2, 3, \ldots \).
Thus, the years in Sequence 2 are:
- \( b_1 = 1996 \)
- \( b_2 = 2005 \)
- \( b_3 = 2014 \)
- \( b_4 = 2023 \)
- \( b_5 = 2032 \), and so on.
Now we need to find the next common year after 2005 in both sequences. The terms in both sequences can be represented in a modular congruence form to find intersections:
For Sequence 1:
\[
a_n = 1991 + 7n
\]
For Sequence 2:
\[
b_n = 1987 + 9m
\]
We want to find the next \( n \) and \( m \) such that:
\[
1991 + 7n = 1987 + 9m
\]
Simplifying this gives:
\[
7n - 9m = -4
\]
Now, we will check when this equation holds after \( n = 2 \) (since 2005 corresponds to \( n = 2 \) in Sequence 1).
To find integer solutions for \( n \) and \( m \), we can explore values of \( n \) starting from \( n = 3 \) (which corresponds to the year 2012).
1. For \( n = 3 \):
\[
7(3) - 9m = -4 \implies 21 - 9m = -4 \implies 9m = 25 \quad (\text{no integer solution})
\]
2. For \( n = 4 \):
\[
7(4) - 9m = -4 \implies 28 - 9m = -4 \implies 9m = 32 \quad (\text{no integer solution})
\]
3. For \( n = 5 \):
\[
7(5) - 9m = -4 \implies 35 - 9m = -4 \implies 9m = 39 \implies m = 4.33 \quad (\text{no integer solution})
\]
4. For \( n = 6 \):
\[
7(6) - 9m = -4 \implies 42 - 9m = -4 \implies 9m = 46 \quad (\text{no integer solution})
\]
5. For \( n = 7 \):
\[
7(7) - 9m = -4 \implies 49 - 9m = -4 \implies 9m = 53 \quad (\text{no integer solution})
\]
6. For \( n = 8 \):
\[
7(8) - 9m = -4 \implies 56 - 9m = -4 \implies 9m = 60 \implies m = \frac{60}{9} \quad (\text{no integer solution})
\]
Continuing this method, we'll realize that we need to find the smallest \( n \) such that:
\[
7n + 4 \equiv 0 \ (\text{mod } 9)
\]
Using \( n = 4 \) (where \( 28 + 4 \equiv 0 \text{ mod } 9\)) works out to give a common year. Therefore the computation yields \( 2012 \).
Finally, checking the years, we find the next year after 2005 that appears in both sequences is:
\[
\text{The next common year is } \boxed{2012}.
\]
**Sequence 1:**
The first term is 1998 and the sequence appears to increase by 7 years each time. The terms can be expressed as:
\[
a_n = 1998 + 7(n - 1) = 1991 + 7n
\]
for \( n = 1, 2, 3, \ldots \).
Thus, the years in Sequence 1 are:
- \( a_1 = 1998 \)
- \( a_2 = 2005 \)
- \( a_3 = 2012 \)
- \( a_4 = 2019 \)
- \( a_5 = 2026 \)
- \( a_6 = 2033 \), and so on.
**Sequence 2:**
The first term is 1996 and the sequence also appears to increase by 9 years each time. The terms can be expressed as:
\[
b_n = 1996 + 9(n - 1) = 1987 + 9n
\]
for \( n = 1, 2, 3, \ldots \).
Thus, the years in Sequence 2 are:
- \( b_1 = 1996 \)
- \( b_2 = 2005 \)
- \( b_3 = 2014 \)
- \( b_4 = 2023 \)
- \( b_5 = 2032 \), and so on.
Now we need to find the next common year after 2005 in both sequences. The terms in both sequences can be represented in a modular congruence form to find intersections:
For Sequence 1:
\[
a_n = 1991 + 7n
\]
For Sequence 2:
\[
b_n = 1987 + 9m
\]
We want to find the next \( n \) and \( m \) such that:
\[
1991 + 7n = 1987 + 9m
\]
Simplifying this gives:
\[
7n - 9m = -4
\]
Now, we will check when this equation holds after \( n = 2 \) (since 2005 corresponds to \( n = 2 \) in Sequence 1).
To find integer solutions for \( n \) and \( m \), we can explore values of \( n \) starting from \( n = 3 \) (which corresponds to the year 2012).
1. For \( n = 3 \):
\[
7(3) - 9m = -4 \implies 21 - 9m = -4 \implies 9m = 25 \quad (\text{no integer solution})
\]
2. For \( n = 4 \):
\[
7(4) - 9m = -4 \implies 28 - 9m = -4 \implies 9m = 32 \quad (\text{no integer solution})
\]
3. For \( n = 5 \):
\[
7(5) - 9m = -4 \implies 35 - 9m = -4 \implies 9m = 39 \implies m = 4.33 \quad (\text{no integer solution})
\]
4. For \( n = 6 \):
\[
7(6) - 9m = -4 \implies 42 - 9m = -4 \implies 9m = 46 \quad (\text{no integer solution})
\]
5. For \( n = 7 \):
\[
7(7) - 9m = -4 \implies 49 - 9m = -4 \implies 9m = 53 \quad (\text{no integer solution})
\]
6. For \( n = 8 \):
\[
7(8) - 9m = -4 \implies 56 - 9m = -4 \implies 9m = 60 \implies m = \frac{60}{9} \quad (\text{no integer solution})
\]
Continuing this method, we'll realize that we need to find the smallest \( n \) such that:
\[
7n + 4 \equiv 0 \ (\text{mod } 9)
\]
Using \( n = 4 \) (where \( 28 + 4 \equiv 0 \text{ mod } 9\)) works out to give a common year. Therefore the computation yields \( 2012 \).
Finally, checking the years, we find the next year after 2005 that appears in both sequences is:
\[
\text{The next common year is } \boxed{2012}.
\]