Question

Let $a_1, a_2, \dots, a_{3n}$ be an arithmetic progression with $a_1 = 3$ and $a_2 = 7$. If $a_1 + a_2 + \dots + a_{3n} = 1830$, then what is the smallest positive integer $m$ such that $m(a_1 + a_2 + \dots + a_n) > 1830$?

Accepted Answer

B. $a_1 = 3, d = 4$. Sum of $3n$ terms $S_{3n} = \frac{3n}{2} [2(3) + (3n - 1)4] = \frac{3n}{2} [12n + 2] = 3n(6n + 1) = 1830$. $n(6n + 1) = 610 \implies 6n^2 + n - 610 = 0 \implies (n - 10)(6n + 61) = 0 \implies n = 10$. Sum of $n$ terms $S_n = S_{10} = \frac{10}{2} [2(3) + 9(4)] = 5(42) = 210$. We want $m S_{10} > 1830 \implies 210m > 1830 \implies m > \frac{1830}{210} = \frac{61}{7} = 8.71$. Smallest integer $m = 9$.

Sum of Terms in Arithmetic Progression

Explanation

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