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Veritas Prep 10 Year Anniversary Promo Question #3

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A sequence is given by the rule a_{n} = |a_{(n-2)}| - |a_{(n-1)}| for all n\geq{3}, where a_1 = 0 and a_2 = 3.

A function s_{n} is defined as the sum of all the terms of the sequence from its beginning through a_n. For instance, s_{4} = a_1 + a_2 + a_3 + a_4. What is s_{101}?

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:

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}. _________________

Ans=C, i have considered first upto a12, and observed that s1=0, s2=3, s3=0 and the cycle repeats.. by this we can divide 101 as 99+2 i.e., s99 would be 0 and s100 =0 and s101=3.

Last edited by chetanachethu on 18 Sep 2012, 10:05, edited 2 times in total.

Considering the rule of Cyclicity,we can conclude that the pattern repeat itself after 3 units i.e a1=0,a2=3,a3=-3 ; again a4=0,a5=3,a6=-3 and so on...

Now 101/3 [as the pattern repeat itself after 3 units as above] gibes a quotient 99 and thus sum of a1+a2+..a99=0 [as every 3 units having sum 0]and a100+a101=|0|+|3|=3

If we use the above formula we get a repetitve series of 3,0,-3 as follows a3 = |a2|- |a1| = 3-0 = 3 a4 = |a3| - |a2| = 3-3 =0 a5 = |a4|-|a3| = 0 - 3 = -3 a6 = |a5| - |a4| = |-3| - 0 = 3-0 =3 a7 = |a6| - |a5| = |3| - |-3| = 3 -3 = 0 And this trend will continue as follows a8 = -3 , a9 = 3 ....and so on until a101 = -3 If we observe there is a pattern emerging starting a3 .If we sum a particular triplet,say a3 + a4+ a5 = 3+0+-3 = 0. Our sum will be 0 Similiarly if we keep adding triplets a6 + a7 + a8 =0 a9 + a10 + a11 = 0 . . . a99 + a100 + a101 = 3 + 0 + -3 = 0. So we can conclude the sum of all the triplets is 0. a3 + .a4 + a5 + .......a101 = 0.

S101 = a1 + a2 + (sum of all the triplets present ) = 0 + 3 + 0 Hence S101 = 3.

A sequence is given by the rule a_{n} = |a_{(n-2)}| - |a_{(n-1)}| for all n\geq{3}, where a_1 = 0 and a_2 = 3.

A function s_{n} is defined as the sum of all the terms of the sequence from its beginning through a_n. For instance, s_{4} = a_1 + a_2 + a_3 + a_4. What is s_{101}?

(A) -3 (B) 0 (C) 3 (D) 201 (E) 303

a3 = |a1| - |a2| a4 = |a2| - |a3| a5 = |a3| - |a4| .. .. .. .. a100 = |a98| - |a99| a101 = |a99| - |a100| ---------------------- Summing together both sides a3+a4+a5+.......+a101 = |a1| - |a100| ( because all other values will be cancelled out in right hand side. For example, |a2| from 1st equation will be cancelled out from |a2| from 2nd equation, and so on.)

So I figured out the sequence first a1=0, a2=3, a3=-3, a4=0, a5=3, a6=-3, a7=0, a8=3, a9=-3, a10=0....... Every third term is a -3 for a3, a6, a9, a12, a15, a18, a21, a, 24....... They are all -3. So s102 = -3 therefore s101 = 3 Another is that the sum of the first 10 is 0 s10= s1 + s2 +s3 + s4....... is 0 so every tenth interval is 0, therefore s100=0 The answer is 3 which is option C

A sequence is given by the rule a_{n} = |a_{(n-2)}| - |a_{(n-1)}| for all n\geq{3}, where a_1 = 0 and a_2 = 3.

A function s_{n} is defined as the sum of all the terms of the sequence from its beginning through a_n. For instance, s_{4} = a_1 + a_2 + a_3 + a_4. What is s_{101}?[/b]

So the series is 0, 3, -3, 0, 3, -3, for a(1), a(2), a(3), a(4), a(5) and a(6) respectively. If you look at the series carefully, S(3) = 0 + 3 + (-3) = 0 this trend will continue and hence S(3), S(6), S(9)..... S(99) = 0 S(101) = S(99) + a(100) + a(101) = 0 + 0 + 3 = 3 S(101) = 3

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:

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}. _________________