Answer :
Sure! Let's solve the problem step-by-step for both cases.
### Case 1: Finding [tex]\( n \)[/tex] such that the [tex]\( n \)[/tex]-th terms of the series [tex]\( 16, 14, 12, \ldots \)[/tex] and [tex]\( 22, 19, 16, \ldots \)[/tex] are equal.
The general form of an arithmetic sequence is:
[tex]\[ T_n = a + (n - 1) \cdot d \][/tex]
For the series [tex]\( 16, 14, 12, \ldots \)[/tex]:
- First term ([tex]\( a_1 \)[/tex]): 16
- Common difference ([tex]\( d_1 \)[/tex]): -2
The [tex]\( n \)[/tex]-th term ([tex]\( T_{n,1} \)[/tex]) is:
[tex]\[ T_{n,1} = 16 + (n - 1) \cdot (-2) \][/tex]
[tex]\[ T_{n,1} = 16 - 2(n - 1) \][/tex]
[tex]\[ T_{n,1} = 16 - 2n + 2 \][/tex]
[tex]\[ T_{n,1} = 18 - 2n \][/tex]
For the series [tex]\( 22, 19, 16, \ldots \)[/tex]:
- First term ([tex]\( a_2 \)[/tex]): 22
- Common difference ([tex]\( d_2 \)[/tex]): -3
The [tex]\( n \)[/tex]-th term ([tex]\( T_{n,2} \)[/tex]) is:
[tex]\[ T_{n,2} = 22 + (n - 1) \cdot (-3) \][/tex]
[tex]\[ T_{n,2} = 22 - 3(n - 1) \][/tex]
[tex]\[ T_{n,2} = 22 - 3n + 3 \][/tex]
[tex]\[ T_{n,2} = 25 - 3n \][/tex]
Setting [tex]\( T_{n,1} \)[/tex] equal to [tex]\( T_{n,2} \)[/tex]:
[tex]\[ 18 - 2n = 25 - 3n \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ 18 - 2n = 25 - 3n \][/tex]
[tex]\[ 18 + n = 25 \][/tex]
[tex]\[ n = 25 - 18 \][/tex]
[tex]\[ n = 7 \][/tex]
So, the [tex]\( n \)[/tex]-th terms of the two series are equal when [tex]\( n = 7 \)[/tex].
### Case 2: Finding [tex]\( n \)[/tex] such that the [tex]\( n \)[/tex]-th terms of the series [tex]\( -9, -6, -3, \ldots \)[/tex] and [tex]\( 16, 14, 12, \ldots \)[/tex] are equal.
For the series [tex]\( -9, -6, -3, \ldots \)[/tex]:
- First term ([tex]\( a_3 \)[/tex]): -9
- Common difference ([tex]\( d_3 \)[/tex]): 3
The [tex]\( n \)[/tex]-th term ([tex]\( T_{n,3} \)[/tex]) is:
[tex]\[ T_{n,3} = -9 + (n - 1) \cdot 3 \][/tex]
[tex]\[ T_{n,3} = -9 + 3(n - 1) \][/tex]
[tex]\[ T_{n,3} = -9 + 3n - 3 \][/tex]
[tex]\[ T_{n,3} = 3n - 12 \][/tex]
For the series [tex]\( 16, 14, 12, \ldots \)[/tex]:
We already derived:
[tex]\[ T_{n,1} = 18 - 2n \][/tex]
Setting [tex]\( T_{n,3} \)[/tex] equal to [tex]\( T_{n,1} \)[/tex]:
[tex]\( 3n - 12 = 18 - 2n \)[/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ 3n - 12 = 18 - 2n \][/tex]
[tex]\[ 3n + 2n = 18 + 12 \][/tex]
[tex]\[ 5n = 30 \][/tex]
[tex]\[ n = \frac{30}{5} = 6 \][/tex]
So, the [tex]\( n \)[/tex]-th terms of the two series are equal when [tex]\( n = 6 \)[/tex].
### Summary
- The [tex]\( n \)[/tex]-th terms of the series [tex]\( 16, 14, 12, \ldots \)[/tex] and [tex]\( 22, 19, 16, \ldots \)[/tex] are equal when [tex]\( n = 7 \)[/tex].
- The [tex]\( n \)[/tex]-th terms of the series [tex]\( -9, -6, -3, \ldots \)[/tex] and [tex]\( 16, 14, 12, \ldots \)[/tex] are equal when [tex]\( n = 6 \)[/tex].
### Case 1: Finding [tex]\( n \)[/tex] such that the [tex]\( n \)[/tex]-th terms of the series [tex]\( 16, 14, 12, \ldots \)[/tex] and [tex]\( 22, 19, 16, \ldots \)[/tex] are equal.
The general form of an arithmetic sequence is:
[tex]\[ T_n = a + (n - 1) \cdot d \][/tex]
For the series [tex]\( 16, 14, 12, \ldots \)[/tex]:
- First term ([tex]\( a_1 \)[/tex]): 16
- Common difference ([tex]\( d_1 \)[/tex]): -2
The [tex]\( n \)[/tex]-th term ([tex]\( T_{n,1} \)[/tex]) is:
[tex]\[ T_{n,1} = 16 + (n - 1) \cdot (-2) \][/tex]
[tex]\[ T_{n,1} = 16 - 2(n - 1) \][/tex]
[tex]\[ T_{n,1} = 16 - 2n + 2 \][/tex]
[tex]\[ T_{n,1} = 18 - 2n \][/tex]
For the series [tex]\( 22, 19, 16, \ldots \)[/tex]:
- First term ([tex]\( a_2 \)[/tex]): 22
- Common difference ([tex]\( d_2 \)[/tex]): -3
The [tex]\( n \)[/tex]-th term ([tex]\( T_{n,2} \)[/tex]) is:
[tex]\[ T_{n,2} = 22 + (n - 1) \cdot (-3) \][/tex]
[tex]\[ T_{n,2} = 22 - 3(n - 1) \][/tex]
[tex]\[ T_{n,2} = 22 - 3n + 3 \][/tex]
[tex]\[ T_{n,2} = 25 - 3n \][/tex]
Setting [tex]\( T_{n,1} \)[/tex] equal to [tex]\( T_{n,2} \)[/tex]:
[tex]\[ 18 - 2n = 25 - 3n \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ 18 - 2n = 25 - 3n \][/tex]
[tex]\[ 18 + n = 25 \][/tex]
[tex]\[ n = 25 - 18 \][/tex]
[tex]\[ n = 7 \][/tex]
So, the [tex]\( n \)[/tex]-th terms of the two series are equal when [tex]\( n = 7 \)[/tex].
### Case 2: Finding [tex]\( n \)[/tex] such that the [tex]\( n \)[/tex]-th terms of the series [tex]\( -9, -6, -3, \ldots \)[/tex] and [tex]\( 16, 14, 12, \ldots \)[/tex] are equal.
For the series [tex]\( -9, -6, -3, \ldots \)[/tex]:
- First term ([tex]\( a_3 \)[/tex]): -9
- Common difference ([tex]\( d_3 \)[/tex]): 3
The [tex]\( n \)[/tex]-th term ([tex]\( T_{n,3} \)[/tex]) is:
[tex]\[ T_{n,3} = -9 + (n - 1) \cdot 3 \][/tex]
[tex]\[ T_{n,3} = -9 + 3(n - 1) \][/tex]
[tex]\[ T_{n,3} = -9 + 3n - 3 \][/tex]
[tex]\[ T_{n,3} = 3n - 12 \][/tex]
For the series [tex]\( 16, 14, 12, \ldots \)[/tex]:
We already derived:
[tex]\[ T_{n,1} = 18 - 2n \][/tex]
Setting [tex]\( T_{n,3} \)[/tex] equal to [tex]\( T_{n,1} \)[/tex]:
[tex]\( 3n - 12 = 18 - 2n \)[/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ 3n - 12 = 18 - 2n \][/tex]
[tex]\[ 3n + 2n = 18 + 12 \][/tex]
[tex]\[ 5n = 30 \][/tex]
[tex]\[ n = \frac{30}{5} = 6 \][/tex]
So, the [tex]\( n \)[/tex]-th terms of the two series are equal when [tex]\( n = 6 \)[/tex].
### Summary
- The [tex]\( n \)[/tex]-th terms of the series [tex]\( 16, 14, 12, \ldots \)[/tex] and [tex]\( 22, 19, 16, \ldots \)[/tex] are equal when [tex]\( n = 7 \)[/tex].
- The [tex]\( n \)[/tex]-th terms of the series [tex]\( -9, -6, -3, \ldots \)[/tex] and [tex]\( 16, 14, 12, \ldots \)[/tex] are equal when [tex]\( n = 6 \)[/tex].
