
Write the first three terms in each of the sequence defined by the following:
1) ${a_n} = 3n + 2$
2) ${a_n} = {n^2} + 1$
Answer
493.5k+ views
Hint: The sequence of each term gives as many different values.
Our discussion for this sequence is about the first three terms only. Yet the sequence \[{a_n}\] has n number of values, when we put different values to the sequence it gives a different number.
Now, we are going to substitute positive integers for n of order \[1,2,3\] that is \[{a_1},{a_2},{a_3}\].
Complete step-by-step answer:
1) \[{a_n} = 3n + 2\]
Take n values for first three terms
So, \[n = 1,2,3\] we get
If we take \[n = 1\] ,
\[ \Rightarrow {a_1} = 3(1) + 2\]
\[ \Rightarrow {a_1} = 3 + 2\]
\[ \Rightarrow {a_1} = 5\]
If we take \[n = 2\] ,
\[ \Rightarrow {a_2} = 3(2) + 2\]
\[ \Rightarrow {a_2} = 6 + 2\]
\[ \Rightarrow {a_2} = 8\]
If we take \[n = 3\] ,
\[ \Rightarrow {a_3} = 3(3) + 2\]
\[ \Rightarrow {a_3} = 9 + 2\]
\[ \Rightarrow {a_3} = 11\]
Hence we get, \[{a_1} = 5,{a_2} = 8,{a_3} = 11\]
The first three terms for the sequence \[{a_n} = 3n + 2\] is \[5,8,11\]
2) \[{a_n} = {n^2} + 1\]
Take n values for first three terms
So, \[n = 1,2,3\]
If we take \[n = 1\]
\[ \Rightarrow {a_1} = {1^2} + 1\]
\[ \Rightarrow {a_1} = 1 + 1\]
\[ \Rightarrow {a_1} = 2\]
If we take \[n = 2\] ,
\[ \Rightarrow {a_2} = {2^2} + 1\]
\[ \Rightarrow {a_2} = 4 + 1\]
\[ \Rightarrow {a_2} = 5\]
If we take \[n = 3\] ,
\[ \Rightarrow {a_3} = {3^2} + 1\]
\[ \Rightarrow {a_3} = 9 + 1\]
\[ \Rightarrow {a_3} = 10\]
Hence we get, \[{a_1} = 2,{a_2} = 5,{a_3} = 10\]
$\therefore $The first three terms for the sequence \[{a_n} = {n^2} + 1\] is \[2,5,10\]
Note:Form the above observation in first sequence \[{a_n} = 3n + 2\]
if n is an odd number, the sequence is odd.
If n is an even number, the sequence is even.
In same way, the sequence \[{a_n} = {n^2} + 1\]
If n is an odd number, the sequence is even.
If n is an even number, the sequence is odd.
Both the sequences have n number of terms, for our convenience we take the first three terms. In some cases, they ask randomly, give value for \[n = 10\] or \[n = 20\] for the sequence, we can find that also by substitution.
Our discussion for this sequence is about the first three terms only. Yet the sequence \[{a_n}\] has n number of values, when we put different values to the sequence it gives a different number.
Now, we are going to substitute positive integers for n of order \[1,2,3\] that is \[{a_1},{a_2},{a_3}\].
Complete step-by-step answer:
1) \[{a_n} = 3n + 2\]
Take n values for first three terms
So, \[n = 1,2,3\] we get
If we take \[n = 1\] ,
\[ \Rightarrow {a_1} = 3(1) + 2\]
\[ \Rightarrow {a_1} = 3 + 2\]
\[ \Rightarrow {a_1} = 5\]
If we take \[n = 2\] ,
\[ \Rightarrow {a_2} = 3(2) + 2\]
\[ \Rightarrow {a_2} = 6 + 2\]
\[ \Rightarrow {a_2} = 8\]
If we take \[n = 3\] ,
\[ \Rightarrow {a_3} = 3(3) + 2\]
\[ \Rightarrow {a_3} = 9 + 2\]
\[ \Rightarrow {a_3} = 11\]
Hence we get, \[{a_1} = 5,{a_2} = 8,{a_3} = 11\]
The first three terms for the sequence \[{a_n} = 3n + 2\] is \[5,8,11\]
2) \[{a_n} = {n^2} + 1\]
Take n values for first three terms
So, \[n = 1,2,3\]
If we take \[n = 1\]
\[ \Rightarrow {a_1} = {1^2} + 1\]
\[ \Rightarrow {a_1} = 1 + 1\]
\[ \Rightarrow {a_1} = 2\]
If we take \[n = 2\] ,
\[ \Rightarrow {a_2} = {2^2} + 1\]
\[ \Rightarrow {a_2} = 4 + 1\]
\[ \Rightarrow {a_2} = 5\]
If we take \[n = 3\] ,
\[ \Rightarrow {a_3} = {3^2} + 1\]
\[ \Rightarrow {a_3} = 9 + 1\]
\[ \Rightarrow {a_3} = 10\]
Hence we get, \[{a_1} = 2,{a_2} = 5,{a_3} = 10\]
$\therefore $The first three terms for the sequence \[{a_n} = {n^2} + 1\] is \[2,5,10\]
Note:Form the above observation in first sequence \[{a_n} = 3n + 2\]
if n is an odd number, the sequence is odd.
If n is an even number, the sequence is even.
In same way, the sequence \[{a_n} = {n^2} + 1\]
If n is an odd number, the sequence is even.
If n is an even number, the sequence is odd.
Both the sequences have n number of terms, for our convenience we take the first three terms. In some cases, they ask randomly, give value for \[n = 10\] or \[n = 20\] for the sequence, we can find that also by substitution.
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