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Question 34 : 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

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Explanatory Answer

Method of solving this CAT Question from Progressions

Let us assume 3n = k
⟹ \\frac{k}{2}) (2a + (k – 1)d) = 1830 we know that a = 3, d = 4
⟹ \\frac{k}{2}) (2(3) + (k – 1)4) = 1830
⟹ \\frac{k}{2}) (6 + 4k – 4) = 1830
⟹ k (2k + 1) = 1830
⟹ 2k2 + k = 1830
⟹ 2k2 + k – 1830 = 0
By factorizing we can find that k = 30, n = 10
⟹ \\frac{10}{2}) (2(3) + 4(9)) = 5(6 + 36)
⟹ 5(42) = 210
⟹ m (a1 + a2 + ..... + an) > 1830
⟹ 210 × m > 1830
⟹ m = 9, since 210 × 9 = 1890

The question is "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 ?"

Hence, the answer is 9

Choice B is the correct answer

 

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