Discussion

Explanation:

a1 = 3, a2 = 7, …..
an = 3 + (n - 1) × 4 = 4n – 1,
a3n = 3 + (3n – 1) × 4 = 12n - 1

a1 + a2 + a3 + … + a3n3n2(12n-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 = 102(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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