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Question: How do you write an explicit formula for the general term of the sequence \([ - 4,2,8,14,...]\)?...

How do you write an explicit formula for the general term of the sequence [4,2,8,14,...][ - 4,2,8,14,...]?

Explanation

Solution

In this question, we need to provide the formula to represent the general term of a sequence. Firstly, observe the sequence given and try to find out which sequence it is. If we see the above sequence, the difference between any term is the same, so it is an arithmetic sequence. We use the formula an=a+(n1)d{a_n} = a + (n - 1)d, where a is the first term and d is the common difference. We substitute the values known and find the required solution.

Complete step by step solution:
Given the sequence of numbers [4,2,8,14,...][ - 4,2,8,14,...]
We are asked to write an explicit formula to find out the general term of the above given sequence.
Generally we represent the sequence by [a1,a2,a3,a4,....][{a_1},{a_2},{a_3},{a_4},....]
Firstly, look at the sequence given. Try to figure out anything common in the given sequence of numbers.
Note that the given sequence looks similar to an arithmetic sequence. To verify this let us find out the common difference d between the any two terms.
To find out common difference d we have the formula d=anan1d = {a_n} - {a_{n - 1}}.
Here a1=4,{a_1} = - 4, a2=2,{a_2} = 2, a3=8,{a_3} = 8, a4=14{a_4} = 14
Now we find out the common difference d.
For n=2n = 2, we have,
d=a2a21d = {a_2} - {a_{2 - 1}}
d=a2a1\Rightarrow d = {a_2} - {a_1}
d=2(4)\Rightarrow d = 2 - ( - 4)
d=2+4\Rightarrow d = 2 + 4
d=6\Rightarrow d = 6
For n=3n = 3, we have,
d=a3a31d = {a_3} - {a_{3 - 1}}
d=a3a2\Rightarrow d = {a_3} - {a_2}
d=82\Rightarrow d = 8 - 2
d=6\Rightarrow d = 6
For n=4n = 4, we have,
d=a4a41d = {a_4} - {a_{4 - 1}}
d=a4a3\Rightarrow d = {a_4} - {a_3}
d=148\Rightarrow d = 14 - 8
d=6\Rightarrow d = 6
Hence the common difference d is constant. Hence the given sequence of numbers is an arithmetic sequence.
The general formula for the arithmetic sequence is given by,
an=a+(n1)d{a_n} = a + (n - 1)d …… (1)
where an{a_n} is the nthn^{th} term of the sequence
aa is the first term and dd is the common difference.
Here a=a1=4,a = {a_1} = - 4, d=6d = 6
Substituting in the equation (1) we get,
an=4+(n1)6{a_n} = - 4 + (n - 1) \cdot 6
an=4+6n6\Rightarrow {a_n} = - 4 + 6n - 6
Combining the like terms we get,
an=46+6n\Rightarrow {a_n} = - 4 - 6 + 6n
an=10+6n\Rightarrow {a_n} = - 10 + 6n
This can also be written as,
an=6n10\Rightarrow {a_n} = 6n - 10

Hence the explicit formula to write the general term of the sequence [4,2,8,14,...][ - 4,2,8,14,...] is given by
an=6n10{a_n} = 6n - 10.

Note: Any sequence in which the difference between every successive term is constant is called the arithmetic sequence.
Students must know how to verify the given list of numbers belongs to the arithmetic sequence. We need to calculate the common difference d which is given by d=anan1d = {a_n} - {a_{n - 1}} between any two terms. If the common difference is constant it is an arithmetic sequence.
If the common difference is not the same then it is not an arithmetic sequence.
They also must know the formula to find the nthn^{th} term which is given by,
an=a+(n1)d{a_n} = a + (n - 1)d
where an{a_n} is the nthn^{th} term of the sequence
aa is the first term and dd is the common difference.