A sequence is given by the rule an = |a_(n-2)| - |a(n-1)|

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 \(s_{101}\).

hi Bunuel

The sum of numbers until a99 will be zero, that's okay, but how can a100 be zero..?

At a99 the cycle is complete, the new cycle begins at a100, then how ..?

please say to me

I have tried this way
99 is a multiple of 3, so when the cycle is complete a99 corresponds to a3, and 100 is multiple of 4, so 100 corresponds to a4 ending at 0, so a101 is equal to 3

please help me work it out

thanks in advance, man

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.

thanks Bunuel

if started from "0", a99 will correspond to "-3" and will sum to "0", then a100 will have value "0" and 101 will be "3"

now it is very clear to me, thank to you again, man

can you please, however, provide me some questions of this kind for further practice?
thanks