
If the \[{n^{th}}\] term of an AP is\[6n + 2\] . Find the \[{9^{th}}\] term.
Answer
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Hint: Sequence is basically a set of things that are in any order. Arithmetic sequence is a sequence where the difference between each successive pair of terms is the same and it is abbreviated as AP. Next term of any sequence can be obtained by adding a constant number to the term before it. That constant number which is added is known as a common difference. Since, all the arithmetic sequences follow the same pattern, we can write the same rule for finding the nth term for the sequence.
Formula used: \[{a_n} = {a_1} + (n - 1)d\]
Complete step by step answer:
We are given,
\[{a_n} = 6n + 2\]
First we need to find $a\;and\;d$ so that we can find any term of the sequence.
Where,
\[
a = first\;term \\
d = common\;difference \\
\]
$
\Rightarrow {a_1} = \;6(1) + 2 = 8 \\
\Rightarrow {a_2} = \;6(2) + 2 = 14 \\
\Rightarrow {a_3} = \;6(3) + 2 = 20 \\
$
Now, we have the first term and we can calculate common differences by subtracting two consecutive terms.
$
\Rightarrow a = 8 \\
\Rightarrow d = 14 - 8 = 6 \\
$
Now we’ll put these values to obtain the required result.
\[ \Rightarrow {a_9} = 8 + (9 - 1)6\]
\[ \Rightarrow {a_9} = 8 + (8)6\]
\[ \Rightarrow {a_9} = 8 + 48\]
\[ \Rightarrow {a_9} = 56\]
This is the required answer.
Alternative Method:
We are given,
\[{a_n} = 6n + 2\]
We have to find the value when $n = 9$ , so we’ll put the value of $n$ in the expression.
\[ \Rightarrow {a_9} = 6(9) + 2\]
\[ \Rightarrow {a_9} = 54 + 2\]
\[ \Rightarrow {a_9} = 56\]
This is the required answer.
Note: Common difference of any sequence can be obtained by subtracting any of the two terms i.e. subtracting the latter term from prior.
$d = {a_n} - {a_{n - 1}}$
Also, Arithmetic Sequence can be both finite and infinite. The behavior of AP depends on the nature of common difference.
If the common difference is a positive number, the sequence will progress towards infinity.
If the common difference is a negative number, the sequence will regress towards negative infinity.
Formula used: \[{a_n} = {a_1} + (n - 1)d\]
Complete step by step answer:
We are given,
\[{a_n} = 6n + 2\]
First we need to find $a\;and\;d$ so that we can find any term of the sequence.
Where,
\[
a = first\;term \\
d = common\;difference \\
\]
$
\Rightarrow {a_1} = \;6(1) + 2 = 8 \\
\Rightarrow {a_2} = \;6(2) + 2 = 14 \\
\Rightarrow {a_3} = \;6(3) + 2 = 20 \\
$
Now, we have the first term and we can calculate common differences by subtracting two consecutive terms.
$
\Rightarrow a = 8 \\
\Rightarrow d = 14 - 8 = 6 \\
$
Now we’ll put these values to obtain the required result.
\[ \Rightarrow {a_9} = 8 + (9 - 1)6\]
\[ \Rightarrow {a_9} = 8 + (8)6\]
\[ \Rightarrow {a_9} = 8 + 48\]
\[ \Rightarrow {a_9} = 56\]
This is the required answer.
Alternative Method:
We are given,
\[{a_n} = 6n + 2\]
We have to find the value when $n = 9$ , so we’ll put the value of $n$ in the expression.
\[ \Rightarrow {a_9} = 6(9) + 2\]
\[ \Rightarrow {a_9} = 54 + 2\]
\[ \Rightarrow {a_9} = 56\]
This is the required answer.
Note: Common difference of any sequence can be obtained by subtracting any of the two terms i.e. subtracting the latter term from prior.
$d = {a_n} - {a_{n - 1}}$
Also, Arithmetic Sequence can be both finite and infinite. The behavior of AP depends on the nature of common difference.
If the common difference is a positive number, the sequence will progress towards infinity.
If the common difference is a negative number, the sequence will regress towards negative infinity.
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