The formula for the sum of the first \( n \) terms of an arithmetic progression (AP) is \( S_n = \frac{n}{2} \left(2a + (n-1)d\right) \).
Given that the sum of the first \( n \) terms is \((n + 2n^2)\), we set the formula equal to this expression:
\[ \frac{n}{2} \left(2a + (n-1)d\right) = n + 2n^2 \]
Multiplying both sides by 2 yields:
\[ n \left(2a + (n-1)d\right) = 2n + 4n^2 \]
Dividing by \( n \) (since \( n eq 0 \)) gives:
\[ 2a + (n-1)d = 2 + 4n \]
The nth term of an AP is \( T_n = a + (n-1)d \).
To find \( T_n \), we rearrange the previous equation:
\[ 2a + (n-1)d = 2 + 4n \]
We can rewrite \( 2a \) as \( a + a \). Therefore, the equation becomes:
\[ a + a + (n-1)d = 2 + 4n \]
Substituting \( T_n = a + (n-1)d \) into this equation, we get:
\[ a + T_n = 2 + 4n \]
This implies \( T_n = 2 + 4n - a \).
Alternatively, we can rewrite \( 2a + (n-1)d \) as \( a + [a + (n-1)d] \), which is \( a + T_n \). So, \( a + T_n = 2 + 4n \).
From \( 2a + (n-1)d = 2 + 4n \), we can express \( a \) in terms of \( T_n \) or vice versa.
Let's use \( 2a + (n-1)d = 2 + 4n \).
We can rewrite this as \( a + (a + (n-1)d) = 2 + 4n \), so \( a + T_n = 2 + 4n \).
To find \( T_n \), we need to determine \( a \).
Let's consider the relationship between \( T_n \) and the given sum.
The first term \( a = T_1 \). The sum of the first term is \( S_1 = n + 2n^2 |_{n=1} = 1 + 2(1)^2 = 3 \). So, \( a = 3 \).
Substituting \( a = 3 \) into \( a + T_n = 2 + 4n \):
\[ 3 + T_n = 2 + 4n \]
\[ T_n = 2 + 4n - 3 \]
\[ T_n = 4n - 1 \]
Let's verify this using the formula \( T_n = a + (n-1)d \).
If \( T_n = 4n - 1 \), then \( T_1 = 4(1) - 1 = 3 \), which matches \( a \).
\( T_2 = 4(2) - 1 = 7 \).
The common difference \( d = T_2 - T_1 = 7 - 3 = 4 \).
Now, let's check the sum formula with \( a=3 \) and \( d=4 \):
\[ S_n = \frac{n}{2} (2(3) + (n-1)4) = \frac{n}{2} (6 + 4n - 4) = \frac{n}{2} (2 + 4n) = n(1 + 2n) = n + 2n^2 \]
This matches the given sum. Therefore, the nth term is \( T_n = 4n - 1 \).
According to the question, this nth term is divisible by 9:
\[ 4n - 1 \equiv 0 \pmod{9} \]
Rearranging gives:
\[ 4n \equiv 1 \pmod{9} \]
To solve for \( n \), multiply both sides by the modular inverse of 4 mod 9. The inverse of 4 mod 9 is 7, since \( 4 \times 7 = 28 \equiv 1 \pmod{9} \).
\[ 7 \times 4n \equiv 7 \times 1 \pmod{9} \]
\[ 28n \equiv 7 \pmod{9} \]
\[ n \equiv 7 \pmod{9} \]
Thus, the smallest possible value of \( n \) is:
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