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A streaming service had 6,000 subscribers at the end of its first month of operation, and the number of subscribers tripled in each month after that. At the end of which month did the service first have 1,458,000 subscribers?
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Correct answer
C
Method (geometric-sequence term formula): the subscriber counts form a geometric sequence with first term a(1) = 6,000 and common ratio r = 3. The count at the end of month n is a(n) = 6,000 × 3⁽ⁿ⁻¹⁾. Note the exponent is (n − 1), not n: at the end of month 1 the exponent is 0, so a(1) = 6,000 × 1 = 6,000, which matches the given starting count.
Set a(n) = 1,458,000: divide both sides by 6,000 to get 3⁽ⁿ⁻¹⁾ = 1,458,000 / 6,000 = 243. Since 243 = 3 × 3 × 3 × 3 × 3 = 3⁵, the exponent (n − 1) = 5, so n = 6. The service first reaches 1,458,000 subscribers at the end of the 6th month. Answer: (C).
Method (direct listing, fastest check): write the monthly counts and stop when you hit the target. Month 1: 6,000. Month 2: 18,000. Month 3: 54,000. Month 4: 162,000. Month 5: 486,000. Month 6: 1,458,000. Counting the listed terms, 1,458,000 is the 6th value, confirming the 6th month.
Why each wrong choice is tempting: (B) the 5th month is the dominant trap: a solver correctly finds that 6,000 is tripled 5 times but reports the number of triplings (the exponent 5) as the month, forgetting that there are 6 terms because the first month is already counted before any tripling. (A) the 4th month comes from numbering the first listed value as month 0 and also dropping the plus-one correction, ending two months early. (D) the 7th month over-corrects by adding the plus-one twice, as if the opening month were itself a tripling step. (E) the 8th month comes from continuing to multiply by 3 past 1,458,000 and reading off a later month. The reliable safeguard is that the term number is always one more than how many times the ratio has been applied.
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