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Question:

Let a₁, a₂,.......a_3n be an arithmetic progression with a₁ = 3 and a₂ = 7. If a₁ + a₂ + ......+ a_3n = 1830, then what is the smallest positive integer m such that m(a₁ + a₂ + ..... + a_n) > 1830?

Explanation:

a₁ = 3, a₂ = 7, …..
a_n = 3 + (n - 1) × 4 = 4n – 1,
a_3n = 3 + (3n – 1) × 4 = 12n - 1

a₁ + a₂ + a₃ + … + a_3n = $\frac{3 n}{2} (12 n - 1 + 3)$ = 1830
⇒ n(6n + 1) = 610
⇒ 6n + n – 610 = 0
⇒ (6n + 61)(n - 10) = 0
⇒ n - 10 = 0

Now a₁₀ = 3 + (10 - 1) × 4 = 39

∴ a₁ + a₂ + a₃ + … + a₁₀ = 3 + 7 + … + 39 = $\frac{10}{2} (3 + 39)$ = 210.

210 × m > 1830

⇒ n > 1830/210 = 8.7.

The minimum integral value of m is 9.

Hence, option (b).

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