in the sequence 2,6,10,... what term has a value of 106?
the difference between consecutive terms is 4
106 = 2 + [4 * (n-1)] ... solve for n , the term number
The sequence that has a last term is called
3
a
Well, if we look at the pattern, it seems to be increasing by 4 each time. So, to find the term that has a value of 106, we can start with 2 and keep adding 4 until we reach 106. Let's see...
2 + 4 = 6
6 + 4 = 10
10 + 4 = 14
14 + 4 = 18
18 + 4 = 22
22 + 4 = 26
26 + 4 = 30
30 + 4 = 34
34 + 4 = 38
38 + 4 = 42
42 + 4 = 46
46 + 4 = 50
50 + 4 = 54
54 + 4 = 58
58 + 4 = 62
62 + 4 = 66
66 + 4 = 70
70 + 4 = 74
74 + 4 = 78
78 + 4 = 82
82 + 4 = 86
86 + 4 = 90
90 + 4 = 94
94 + 4 = 98
98 + 4 = 102
102 + 4 = 106
Ah, it looks like the term that has a value of 106 is the 27th term!
To find the term in the sequence with a value of 106, we first need to determine the pattern of the sequence.
Looking at the given sequence 2, 6, 10, ..., we can notice that each term is increasing by 4. This means that the common difference between the terms is 4.
To find the nth term of an arithmetic sequence, we can use the formula:
nth term = first term + (n - 1) * common difference
In this case, the first term (a₁) is 2, the common difference (d) is 4, and we want to find the term that has a value of 106. We can substitute these values into the formula and solve for n:
106 = 2 + (n - 1) * 4
Now we can solve for n:
104 = (n - 1) * 4
104/4 = n - 1
26 = n - 1
Adding 1 to both sides, we get:
n = 27
Therefore, the 27th term of the sequence has a value of 106.