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

We need to find a pattern for the arithmetic progression and then determine when its nth term is divisible by 9.

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We're told that the sum of the first n terms of an arithmetic progression is:

$S_n = n + 2n^2$

Key Insight: Instead of calculating terms one by one, we can use the relationship between consecutive sums to find a general formula for the nth term.

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For any sequence, we know that:

First term: $a_1 = S_1$

For $n \geq 2$: $a_n = S_n - S_{n-1}$

Let us find the first term:

$a_1 = S_1 = 1 + 2(1)^2 = 1 + 2 = 3$

For $n \geq 2$:

$a_n = S_n - S_{n-1}$

$a_n = (n + 2n^2) - ((n-1) + 2(n-1)^2)$

Let us expand $(n-1)^2 = n^2 - 2n + 1$:

$a_n = n + 2n^2 - (n-1) - 2(n^2 - 2n + 1)$

$a_n = n + 2n^2 - n + 1 - 2n^2 + 4n - 2$

$a_n = 4n - 1$

Let us check if this formula works for $n = 1$:

$a_1 = 4(1) - 1 = 3$

So the general term is: $a_n = 4n - 1$

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We need $a_n$ to be divisible by 9, which means:

$4n - 1 \equiv 0 \pmod{9}$

$4n \equiv 1 \pmod{9}$

We need to find what value of $n$ makes $4n$ leave remainder 1 when divided by 9.

Let us check systematically:

$4 \times 1 = 4$ (remainder 4)

$4 \times 2 = 8$ (remainder 8)

$4 \times 3 = 12$ (remainder 3)

$4 \times 4 = 16$ (remainder 7)

$4 \times 5 = 20$ (remainder 2)

$4 \times 6 = 24$ (remainder 6)

$4 \times 7 = 28$ (remainder 1)

So $n = 7$ is our answer.

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Let us verify that $a_7$ is indeed divisible by 9:

$a_7 = 4(7) - 1 = 28 - 1 = 27$

Since $27 = 3 \times 9$, yes, $a_7$ is divisible by 9.

Why this approach is faster: Instead of calculating 7 different terms individually, we found the general formula and used modular arithmetic to solve directly.

Therefore, the smallest possible value of $n$ is $\boxed{7}$.