At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
