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**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.

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