How do you write the first six terms of the sequence?
Answer
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Hint: Here, we will find the terms in a sequence by using the given \[{n^{th}}\] term formula of an AP i.e. the given equation. Then we will substitute different values of \[n\], to find the required consecutive terms. An arithmetic sequence is a sequence of numbers such that the common difference between any two consecutive numbers is a constant.
Complete Step by Step Solution:
The equation is the \[{n^{th}}\] term formula of an AP.
First, we will find the first term of the sequence by substituting \[n = 1\] in \[{a_n} = 6 - n\]. Therefore, we get
\[{a_1} = 6 - 1\]
Subtracting the terms, we get
\[ \Rightarrow {a_1} = 5\]
Now, we will find the second term of the sequence by substituting \[n = 2\] in \[{a_n} = 6 - n\], so we get
\[{a_2} = 6 - 2\]
Subtracting the terms, we get
\[ \Rightarrow {a_2} = 4\]
We will now find the third term of the sequence by substituting \[n = 3\] in \[{a_n} = 6 - n\]. Therefore, we get
\[{a_3} = 6 - 3\]
Subtracting the terms, we get
\[ \Rightarrow {a_3} = 3\]
Now, we will find the fourth term of the sequence by substituting \[n = 4\] in \[{a_n} = 6 - n\]. So , we get
\[{a_4} = 6 - 4\]
Subtracting the terms, we get
\[ \Rightarrow {a_4} = 2\]
We will now find the fifth term of the sequence by substituting \[n = 5\] in \[{a_n} = 6 - n\] to get
\[{a_5} = 6 - 5\]
Subtracting the terms, we get
\[ \Rightarrow {a_5} = 1\]
Now, we will find the sixth term of the sequence by substituting \[n = 6\] in \[{a_n} = 6 - n\]. Therefore, we get
\[{a_6} = 6 - 6\]
Subtracting the terms, we get
\[ \Rightarrow {a_6} = 0\]
Therefore, the first six terms of the sequence \[{a_n} = 6 - n\] are \[5, 4, 3, 2, 1, 0\].
Note:
We know that a sequence of real numbers is defined as an arrangement or a list of real numbers in a specific order. We should know that if a sequence has only a finite number of terms then it is called a finite sequence and if a sequence has infinitely many terms, then it is called an infinite sequence. If we are given a general term of a sequence and then we will be able to find any particular term of the sequence directly.
Complete Step by Step Solution:
The equation is the \[{n^{th}}\] term formula of an AP.
First, we will find the first term of the sequence by substituting \[n = 1\] in \[{a_n} = 6 - n\]. Therefore, we get
\[{a_1} = 6 - 1\]
Subtracting the terms, we get
\[ \Rightarrow {a_1} = 5\]
Now, we will find the second term of the sequence by substituting \[n = 2\] in \[{a_n} = 6 - n\], so we get
\[{a_2} = 6 - 2\]
Subtracting the terms, we get
\[ \Rightarrow {a_2} = 4\]
We will now find the third term of the sequence by substituting \[n = 3\] in \[{a_n} = 6 - n\]. Therefore, we get
\[{a_3} = 6 - 3\]
Subtracting the terms, we get
\[ \Rightarrow {a_3} = 3\]
Now, we will find the fourth term of the sequence by substituting \[n = 4\] in \[{a_n} = 6 - n\]. So , we get
\[{a_4} = 6 - 4\]
Subtracting the terms, we get
\[ \Rightarrow {a_4} = 2\]
We will now find the fifth term of the sequence by substituting \[n = 5\] in \[{a_n} = 6 - n\] to get
\[{a_5} = 6 - 5\]
Subtracting the terms, we get
\[ \Rightarrow {a_5} = 1\]
Now, we will find the sixth term of the sequence by substituting \[n = 6\] in \[{a_n} = 6 - n\]. Therefore, we get
\[{a_6} = 6 - 6\]
Subtracting the terms, we get
\[ \Rightarrow {a_6} = 0\]
Therefore, the first six terms of the sequence \[{a_n} = 6 - n\] are \[5, 4, 3, 2, 1, 0\].
Note:
We know that a sequence of real numbers is defined as an arrangement or a list of real numbers in a specific order. We should know that if a sequence has only a finite number of terms then it is called a finite sequence and if a sequence has infinitely many terms, then it is called an infinite sequence. If we are given a general term of a sequence and then we will be able to find any particular term of the sequence directly.
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