For any natural number $$ n $$, suppose the sum of the first $$ n $$ terms of an arithmetic progression is
$$ S_n = n + 2n^2 $$.
If the $$ n^ ext{th} $$ term of the progression is divisible by 9, then the smallest possible value of $$ n $$ is:

Question

For any natural number $$ n $$, suppose the sum of the first $$ n $$ terms of an arithmetic progression is  
$$ S_n = n + 2n^2 $$.  
If the $$ n^	ext{th} $$ term of the progression is divisible by 9, then the smallest possible value of $$ n $$ is:

Accepted Answer

To solve this problem:

1. **Sum Formula**:  
   The sum of the first $$ n $$ terms is given as:  
   $$
   S_n = n + 2n^2
   $$

2. **Find the $$ n^	ext{th} $$ Term**:  
   The $$ n^	ext{th} $$ term is calculated as:  
   $$
   T_n = S_n - S_{n-1}
   $$  
   Substitute values:  
   $$
   T_n = (n + 2n^2) - \left((n-1) + 2(n-1)^2ight)
   $$

3. **Simplify Expression for $$ T_n $$**:  
   Expand and simplify:  
   $$
   T_n = n + 2n^2 - \left(n - 1 + 2(n^2 - 2n + 1)ight)
   $$  
   $$
   T_n = n + 2n^2 - (n - 1 + 2n^2 - 4n + 2)
   $$  
   $$
   T_n = n + 2n^2 - n + 1 - 2n^2 + 4n - 2
   $$  
   $$
   T_n = 4n - 1
   $$

4. **Divisibility Condition**:  
   For the $$ n^	ext{th} $$ term to be divisible by 9:  
   $$
   4n - 1 \equiv 0 \, (	ext{mod} \, 9)
   $$

5. **Solve for $$ n $$**:  
   Rearrange:  
   $$
   4n \equiv 1 \, (	ext{mod} \, 9)
   $$  
   Multiply both sides by the modular inverse of 4 modulo 9 (which is 7):  
   $$
   n \equiv (7 	imes 1) \, (	ext{mod} \, 9) \implies n \equiv 7
   $$

Thus, the smallest value of $$ n $$ is **7**.

Answer

Answer

Answer