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

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Re: A sequence is given by the rule an = |a_(n-2)| - |a(n-1)| [#permalink]
Bunuel wrote:

Winner:

yogeshwar007

Official Explanation:

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 ..?

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

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Re: A sequence is given by the rule an = |a_(n-2)| - |a(n-1)| [#permalink]
gmatcracker2017 wrote:
Bunuel wrote:

Winner:

yogeshwar007

Official Explanation:

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 ..?

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

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.
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Re: A sequence is given by the rule an = |a_(n-2)| - |a(n-1)| [#permalink]
Bunuel wrote:
gmatcracker2017 wrote:
Bunuel wrote:

Winner:

yogeshwar007

Official Explanation:

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 ..?

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

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
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Re: A sequence is given by the rule an = |a_(n-2)| - |a(n-1)| [#permalink]
gmatcracker2017 wrote:
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

12. Sequences

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Re: A sequence is given by the rule an = |a_(n-2)| - |a(n-1)| [#permalink]
Bunuel wrote:
gmatcracker2017 wrote:
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

12. Sequences

thanks a lot bunu
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Re: A sequence is given by the rule an = |a_(n-2)| - |a(n-1)| [#permalink]
To solve for s101s101​, the sum of the first 101 terms of the sequence defined by an=∣an−2∣−∣an−1∣an​=∣an−2​∣−∣an−1​∣ with initial conditions a1=0a1​=0 and a2=3a2​=3, we need to compute the sequence up to a101a101​ and then find the cumulative sum.

Let's calculate the sequence step by step:

a1=0a1​=0
a2=3a2​=3
a3=∣a1∣−∣a2∣=∣0∣−∣3∣=0−3=−3a3​=∣a1​∣−∣a2​∣=∣0∣−∣3∣=0−3=−3
a4=∣a2∣−∣a3∣=∣3∣−∣−3∣=3−3=0a4​=∣a2​∣−∣a3​∣=∣3∣−∣−3∣=3−3=0
a5=∣a3∣−∣a4∣=∣−3∣−∣0∣=3−0=3a5​=∣a3​∣−∣a4​∣=∣−3∣−∣0∣=3−0=3
a6=∣a4∣−∣a5∣=∣0∣−∣3∣=0−3=−3a6​=∣a4​∣−∣a5​∣=∣0∣−∣3∣=0−3=−3

Continuing this process until a101a101​:

The sequence alternates between 0, 3, -3 with a repeating cycle every 3 terms: 0,3,−3,0,3,−3,…0,3,−3,0,3,−3,….

To find s101s101​, we need to calculate how many complete cycles of 3 terms fit into 101 terms:

There are ⌊1013⌋=33⌊3101​⌋=33 complete cycles.
In each cycle, the sum 0+3−3=00+3−3=0.

Therefore, the sum of these 33 cycles is 33×0=033×0=0.

Additionally, we need to include any remaining terms beyond these complete cycles:

There are 101−33×3=2101−33×3=2 remaining terms.

The terms a100a100​ and a101a101​ correspond to the sequence elements where the cycle restarts, which are −3−3 and 00 respectively.

So, s101=0+(−3)+0=−3s101​=0+(−3)+0=−3.

Therefore, s101s101​ is −3−3​.­
Re: A sequence is given by the rule an = |a_(n-2)| - |a(n-1)| [#permalink]
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