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

  1. $8$
  2. $9$
  3. $10$
  4. $11$
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Given that, $a_{1},a_{2},a_{3}, \ldots,  a_{3n} $ in AP.

And   $a_{1} = 3, a_{2} = 7$

Common difference $(d) =  a_{2} – a_{1}  = 7-3 = 4$

The sum of $n$term in AP: $S_{n} = \frac{n}{2} [2a + (n-1)d ]$

Where, $a = $first term, $n= $number of terms, $d= $ common difference  

$a_{1} + a_{2} + a_{3} + \ldots + a_{3n} = 1830 $

$\Rightarrow \frac{3n}{2} [ \, 2(3) + (3n-1)4  ]  = 1830$

$\Rightarrow \frac{3n}{2} [ \, 6 + 12n – 4 ]  = 1830$

$\Rightarrow \frac{3n}{2} [ \, 12n + 2 ]  = 1830$

$\Rightarrow n(6n + 1)  = 610$

$\Rightarrow 6n^{2} + n -610$

$\Rightarrow n = \frac {-1\pm \sqrt{1-4(6)(-610)}}{12}$

$\Rightarrow n = \frac {-1\pm \sqrt{1+14640}}{12}$

$\Rightarrow n = \frac {-1\pm \sqrt{14641}}{12}$

$\Rightarrow n = \frac {-1\pm 121}{12}$

$\Rightarrow n = \frac {-1-121}{12}  =  \frac {-122}{12}=  \frac {-61}{6}\quad (\text{The number of terms can't be negative})$

$\Rightarrow n = \frac {-1+121}{12}  =  \frac {120}{12}=  10$

$\Rightarrow \boxed {n = 10}$

Now, $a_{n} = a + (n-1)d$

$\Rightarrow a_{10} = 3 + (10-1)4$   

$\Rightarrow \boxed { a_{10} = 39 }$

Now, $m( a_{1} + a_{2} +  \ldots + a_{n} ) > 1830$

$\Rightarrow m( a_{1} + a_{2} + \ldots + a_{10} ) > 1830$

$\Rightarrow m( 3 + 7 + \ldots  + 39) > 1830 $

$\Rightarrow m\times \frac{10}{2} [ \, 2(3) + (10-1)4  ] > 1830$

$\Rightarrow \frac{10m}{2}  (6 + 36) > 1830$

$\Rightarrow \frac{420m}{2} > 1830$

$\Rightarrow 210m> 1830$

$\Rightarrow m> \frac{183}{21}$

$\Rightarrow m> 8.71$

$\Rightarrow m = \lceil 8.71 \rceil$

$\Rightarrow \boxed { m_{\text{min}} = 9 }$

$\therefore$ The smallest positive integer $m = 9.$

Correct Answer $: \text{B}$
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