Sequences & Series
Sequences
Grade 11
Question:
<p><strong>For Problems 13–15:</strong> Consider the sequence in the form of groups \((1), (2, 2), (3, 3, 3), (4, 4, 4, 4), (5, 5, 5, 5, 5), \ldots\)</p><p>The 2000<sup>th</sup> term of the sequence is not divisible by</p>
<p>3</p>
<p>9</p>
<p>7</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Find which group contains the 2000th term by solving n(n+1)/2 ≥ 2000, then determine the value and its divisibility properties.
<p><strong>Step 1:</strong> Find which group contains the 2000th term. Group n contains n terms, so the total terms through group n is 1 + 2 + 3 + ... + n = n(n+1)/2.</p><p><strong>Step 2:</strong> Solve n(n+1)/2 ≥ 2000, which gives n(n+1) ≥ 4000. Testing: 62(63) = 3906 < 4000 and 63(64) = 4032 > 4000. So the 2000th term is in group 63.</p><p><strong>Step 3:</strong> Terms through group 62: 62(63)/2 = 1953. The 2000th term is the (2000 - 1953) = 47th term in group 63. Since group 63 consists of 63 identical copies of 63, the 2000th term = 63.</p><p><strong>Step 4:</strong> Find divisors of 63 = 3² × 7. The divisors are 1, 3, 7, 9, 21, 63. Therefore 63 is divisible by 1, 3, 7, 9, 21, and 63. The answer is the number NOT in this list.</p><p>∴ Answer: C</p>
Correct Answer: C