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CAT 2017 Slot 1 — QA Question 34

Arithmetic ProgressionEasy

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?

Answer & solution

  • A

    8

  • 9

  • C

    10

  • D

    11

Solution

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

a1 + a2 + a3 + … + a3n = 3n2(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).

CAT 2017 Slot 1 QA Q34: Let a 1 , a 2 ,.......a 3n be an arithmetic progression with a 1 = 3 and a 2 = 7. If a 1 + a 2 + ......+ a 3n — Solution | TheCATExam