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Let a1 , a2 ,........a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ....+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + .... + an ) > 1830?
1. 8
2. 9
3. 10
4. 11

1 Answer

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Correct Answer - Option 2 : 9

Calculation:

⇒ For 3n terms, n = 3n, a = 3, and d = a2 - a1 = 7 - 3 = 4

⇒ (3n/2) × [2 × 3 + (3n - 1) × 4]

⇒ 6n2 + n - 610 = 0

By solving this quadratic equation

⇒ n = 10, and n = -(61/6), n can not be negative we will consider n = 10

⇒ for positive integer m

⇒ m × (3 + 7 + 11 + ............. + 39) > 1830

⇒ m × (10/2) × [2 × 3 + (10 - 1) × 4] > 1830

By solving

⇒ m > 8.71 ≈ 9

∴ The required result will be 9.

Sn = (n/2) × [2a + (n - 1) × d]

Where a = First term, n = Number of term, d = difference

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