1. Solve for the 101st term of the sequence whose 1st term is x-y and d=2x+y-3.
2. In the sequence 2, 6, 10, … , what term has a value of 106?
3. Solve for the 101st term of the sequence: 6-p, 4+p, 2+3p, 5p, … .
In all 3 questions you are using the same formula
term(n) = a + (n-1)d
1. term(101) = x-y + 100(2x+y-3)
= x-y+200x + 100y - 300
= 201x + 100y - 300
2.
a=2 , d=4
2 + (n-1)(4) = 106
4n-4 = 104
4n = 108
n = 27
It is the 27th term
3. give this one a try yourself
x2(4x-8)-12
1. To solve for the 101st term of a sequence, we need to understand the formula for the nth term of an arithmetic sequence. The formula is given as:
a_n = a_1 + (n-1)d
Where:
a_n is the nth term of the sequence
a_1 is the first term of the sequence
n is the position of the term we want to find
d is the common difference between consecutive terms
In this case, the first term is x-y and the common difference is 2x+y-3. So we can plug these values into the formula to find the 101st term:
a_101 = (x-y) + (101-1)(2x+y-3)
= (x-y) + 100(2x+y-3)
= (x-y) + 200x + 100y - 300
Therefore, the 101st term of the sequence is (x-y) + 200x + 100y - 300.
2. In the given sequence 2, 6, 10, …, we can observe that each term is increasing by 4. So the common difference, d, is 4.
We can use the formula for the nth term of an arithmetic sequence to find the term that has a value of 106. Rearranging the formula, we have:
n = (a_n - a_1) / d
Substituting the given values, we get:
n = (106 - 2) / 4
n = 104 / 4
n = 26
Therefore, the term that has a value of 106 is the 26th term.
3. In the given sequence 6-p, 4+p, 2+3p, 5p, …, we can observe that each term has a pattern. To find the 101st term, we need to identify the pattern and use it to generate the desired term.
The pattern in this sequence can be explained as follows:
- The first term, 6-p, has a difference of (-p).
- The second term, 4+p, has a difference of (2p) from the previous term.
- The third term, 2+3p, has a difference of (4p) from the previous term.
- The fourth term, 5p, has a difference of (p) from the previous term.
From this pattern, we can see that the difference alternates between (-p) and (2p) for each term. So, for the 5th term, the difference will be (-2p), and for the 6th term, it will be (3p). This pattern continues.
To find the 101st term, we need to determine whether the term is even or odd. Since 101 is an odd number, the difference will be (2p).
Now, we can use the formula for the nth term of an arithmetic sequence, which is the same as in question 1:
a_n = a_1 + (n-1)d
The first term is 6-p, and the common difference is 2p. So we can substitute the values into the formula:
a_101 = (6-p) + (101-1)(2p)
= (6-p) + 100(2p)
= (6-p) + 200p
Therefore, the 101st term of the sequence is (6-p) + 200p.