In a certain sequence of 10 positive numbers, each number after the first is the square root of the previous number. If the first number is 9, how many numbers are less than 1?
Answer Choices
Correct answer marked belowSee the full step-by-step explanation
You can see the correct answer above. Sign in for free to unlock the complete worked solution.
Track your performance and improve
Get detailed analytics, unlock full explanations, and move up difficulty tiers as you practice.
Unlock the full explanation
Create a free account to reveal the correct answer, see the step-by-step explanation, and start tracking your GMAT progress.
Wait, so each term is the square root of the previous? That means the sequence is 9, 3, sqrt(3), ... I'm getting that from the first term 9, the second is 3, third is sqrt(3) ~1.732, fourth is sqrt(1.732) ~1.316, fifth is sqrt(1.316) ~1.147, sixth is sqrt(1.147) ~1.071, seventh is sqrt(1.071) ~1.035, eighth is sqrt(1.035) ~1.017, ninth is sqrt(1.017) ~1.008, tenth is sqrt(1.008) ~1.004. So none are less than 1? But that seems too easy. Did I mess up the order?
No, you're right! The numbers keep getting closer to 1 but never actually go below 1. So the answer is 0.
But wait, the question says 'how many numbers are less than 1?' If none are less than 1, then 0. But is there a trick? Like does the sequence eventually go below 1 if we had more terms? With 10 terms, it's still above 1.
This one is straightforward if you know that square rooting a number greater than 1 keeps it greater than 1. So all 10 numbers are >1. Answer is 0.
Yeah, but the first number is 9, so all subsequent are positive and >1. So none less than 1.
I initially thought it might drop below 1 after a few terms, but after computing a few, I see it approaches 1 from above. So 0 is correct. The wording 'less than 1' is key; if it said 'less than or equal to 1', still 0 because it never equals 1 either.
This is a classic sequence problem. Since the first term is 9, and each term is the square root of the previous, all terms are positive and greater than 1. So the answer is (A) 0.
But careful: the sequence is 10 positive numbers. So the first is 9, second 3, third sqrt(3) which is about 1.732, which is >1. So indeed none are less than 1.
What it tests
Your ability to spot and extend arithmetic and geometric sequences and work with recursive definitions.
Common trap
Off-by-one errors in indexing (a₀ vs a₁) or confusing an arithmetic difference with a geometric ratio.