Winner:

yogeshwar007

Official Explanation:

Answer is C

While we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “pre-previous” one.

So \(a_3 = |a_1| - |a_2|\), and \(a_4 = |a_2| - |a_3|\), and \(a_5 = |a_3| - |a_4|\), and so on. We are told that the sequence begins 0, 3, ..., so we can derive that:

\(a_3 = |a_1| - |a_2| = |0| - |3| = 0 - 3 = -3\).
Then \(a_4 = |a_2| - |a_3| = |3| - |-3| = 3 - 3 = 0\).
Then \(a_5 = |a_3| - |a_4| = |-3| - |0| = 3 - 0 = 3\).

As soon as we’ve seen \(a_4\) and \(a_5\) turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as \(a_1\) and \(a_2\)) — so \(a_6\) will be identical to \(a_3\), so -3. So we have established the pattern 0, 3, -3, 0, 3, -3, ... for our sequence.

Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + -3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with \(a_{99}\). \(a_{100}\) will then add itself (0) to that sum, and \(a_{101}\) will add itself (3) onto that. So we will land at a sum of 3 for \(s_{101}\).

It's very easy. The sequence is:

{0, 3, -3} {0, 3, -3} {0, 3, -3} {0, 3, -3} ...

Notice that if n is a multiple of 3, so a3, a6, ..., a99, then it's -3. The next terms, so a4, a7, ..., a100 are 0.