Let $a_{1}, a_{2}, \ldots \ldots . . . a_{3 n}$ be an arithmetic progression with $a_{1}=3$ and $a_{2}=7$. If $a_{1}+a_{2}+\ldots .+a_{3 n}=1830$, then what is the smallest positive integer $m$ such that $m\left(a_{1}+a_{2}+\ldots .+a_{n}\right)>1830$ ?

We're given an arithmetic progression problem, and we'll solve it step by step, making sure to explain each concept clearly.

We're given:

$a_1 = 3$ (first term)

$a_2 = 7$ (second term)

In an arithmetic progression, each term differs from the previous by a constant value called the common difference.

Common difference: $d = a_2 - a_1 = 7 - 3 = 4$

General term formula: $a_n = a_1 + (n-1) \times d = 3 + (n-1) \times 4 = 4n - 1$

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We need to use the sum formula for arithmetic progressions. For the first $3n$ terms:

Sum formula: $S_{3n} = \tfrac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})$

Number of terms = $3n$

First term = $a_1 = 3$

Last term = $a_{3n} = 4(3n) - 1 = 12n - 1$

$S_{3n} = \frac{3n}{2} \times (3 + 12n - 1) = \frac{3n}{2} \times (2 + 12n)$

$S_{3n} = \frac{3n(2 + 12n)}{2} = \frac{3n(12n + 2)}{2}$

Since we're told this sum equals 1830:

$\frac{3n(12n + 2)}{2} = 1830$

$3n(12n + 2) = 3660$

$n(12n + 2) = 1220$

$12n^2 + 2n = 1220$

Wait, let us recalculate this more carefully:

$3n(12n + 2) = 3660$

$3n \times 2(6n + 1) = 3660$

$6n(6n + 1) = 3660$

$n(6n + 1) = 610$

$6n^2 + n = 610$

$6n^2 + n - 610 = 0$

$(6n + 61)(n - 10) = 0$

Since $n$ must be positive: $n = 10$

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Now we need: $a_1 + a_2 + \ldots + a_n$ where $n = 10$

$S_n = \frac{n}{2}(a_1 + a_n) = \frac{10}{2}(3 + [4(10) - 1])$

$S_n = 5(3 + 39) = 5 \times 42 = 210$

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We need the smallest positive integer $m$ such that:

$m(a_1 + a_2 + \ldots + a_n) > 1830$

$m \times 210 > 1830$

$m > \frac{1830}{210} = 8.714...$

Since $m$ must be a positive integer, the smallest value is $m = 9$.

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The key insight is that we're comparing the sum of the first $n$ terms with the sum of the first $3n$ terms. Since $3n = 30$ terms sum to 1830, we need to find how many times the sum of the first 10 terms we need to exceed 1830.

Answer: $m = 9$