Problem Solving

As advertised on the main page of Beat the GMAT, Brian, David and I will be posting some original questions today and tomorrow with prizes for the first five correct responders who show their work!

Good Luck!


A sequence is given by the rule an = |an-2| - |an-1| for all n>2, and a function s(n) is defined as the sum of all the terms of this sequence from its beginning through an. For instance, s(4) = a1 + a2 + a3 + a4. If a1 = 0 and a2 = 3, what is the value of s(101)?


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

A sequence is given by the rule an = |an-2| - |an-1| for all n>2, and a function s(n) is defined as the sum of all the terms of this sequence from its beginning through an. For instance, s(4) = a1 + a2 + a3 + a4. If a1 = 0 and a2 = 3, what is the value of s(101)?


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

Answer = C = 3

a1 = 0
a2 = 3
a3 = |a1| - |a2| = -3
a4 = |a2| - |a3| = 0
and this sequence continues

S101 = a1 + a2 + a3+....+a101 --> we know that S3= 0 Hence this translates to S101= S99 + a100 + a101 = 0 + 0 + 3= 3

a3=|a1|-|a2|= |0|-|3| = -3
a4=|a2|-|a3|=|3|-|-3|=3-3=0
S(4)= 0+3-3+0=0
a5=|a3|-|a4|=|-3|-|0|=3
a6=|a4|-|a5|=0-3=-3
a7=3-3=0
a8=3-0=3
a9=-3
a10=0
s(10)=0
So the values follow the seq -3,0,3
S(10)=0 => S(100) = 0
S(101) = 3

a1 = 0
a2 = 3
a3 = |a1|-|a2| = -3
a4 = |a2|-|a3| = 0

the sequence goes = 0,positive,negative...0,positive,negative...
hence, till a99, the sum would be 0
a100 = 0
a101 = 3
a100+a101 = 0 + 3 = 3
Ans : C

the entire an repeats itself in cycle of 0,3,-3

So S3=S6=...=S99=0

a100=0
a101=3

Therefore, S101=s99+a100+a101= 3

So answer is C

Ashley@VeritasPrep wrote:As advertised on the main page of Beat the GMAT, Brian, David and I will be posting some original questions today and tomorrow with prizes for the first five correct responders who show their work!

Good Luck!


A sequence is given by the rule an = |an-2| - |an-1| for all n>2, and a function s(n) is defined as the sum of all the terms of this sequence from its beginning through an. For instance, s(4) = a1 + a2 + a3 + a4. If a1 = 0 and a2 = 3, what is the value of s(101)?


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

Answer is C.

a1 = 0
a2 = 3
a3 = |a1| - |a2| = -3
a4 = |a2| - |a3| = 0
a5 = |a3| - |a4| = 3
a6 = |a4| - |a5| = -3
the pattern of 0, 3, and -3 continues.

Therefore, S101 = 33 (a1 + a2 + a3)+a100+a101 = 33(0)+0+3 = 3

Here, we have
a1 = 0
a2 = 3
Therefore,
a3=|a1|-|a2|= |0|-|3| = -3
a4=|a2|-|a3|=|3|-|-3|=3-3=0
So the values follow the sequence -3,0,3
and this sequence continues....
Therefore Sum of a1+a2+...a101 - > 3

Hence the answer is C

A sequence is given by the rule an = |an-2| - |an-1| for all n>2, and a function s(n) is defined as the sum of all the terms of this sequence from its beginning through an. For instance, s(4) = a1 + a2 + a3 + a4. If a1 = 0 and a2 = 3, what is the value of s(101)?

Solution:
a1 = 0
a2 = 3
a3 = |a1| - |a2| = 0 - 3 = -3
a4 = |a2| - |a3| = 3 - 3 = 0
a5 = |a3| - |a4| = 3 - 0 = 3
a6 = |a4| - |a5| = 0 - 3 = -3

It is in cycle of (0, 3, -3) and sum of each of them is zero
S(101) = a101 + a100 + 0 = 0 + 3 + 0 = 3

Answer = C

Wow! Good Question! I will not pretend that i knew the solution (until i check out other replies!) got stumped at first! Took me a while to understand the Q and A

I am guessing this is a 700-800 Q?

Thanks

melguy wrote:Wow! Good Question! I will not pretend that i knew the solution (until i check out other replies!) got stumped at first! Took me a while to understand the Q and A

I am guessing this is a 700-800 Q?

Thanks

Hey melguy,

Thanks No real stats on the score range for this question, since it's just an original one, but I definitely did intend it to be a challenge, so yes, I'd estimate that the people who figure this one out are looking at some pretty high quant scores!

Ashley@VeritasPrep wrote: Hey melguy,

Thanks No real stats on the score range for this question, since it's just an original one, but I definitely did intend it to be a challenge, so yes, I'd estimate that the people who figure this one out are looking at some pretty high quant scores!

Agree with you! I figured it out but in 5 mins which is equal to guessing on a real test My score is still very low - Q35.

I will look forward to more Q's from you.

Thanks

a1=0,a2=3,a3=3,a4=0,a5=3....a10=0.

thus a100 = 0 as a1-a10 will repeat.
hence a101 = a2=3.

The answer is C (3).

Fitting a1=0 and a2=3 into the formula, I will end up with a series with certain pattern like this :
(0,3,-3),(0,3,-3),(0,3,-3),(0,3,-3), ...
In other words, I will have a repeating group that consists of three elements, and I know that the sum of these elements is 0 (zero).
In order to get the S(101), I need to find the remainder of 101 divided by 3. As 101 = 33x3 + 2, I know that the remainder is 2.

Thus, S(101) = 33 x the sum of the elements in the group (which is zero) + the sum of the next two elements (which is 0 and 3).
In other words, S(101) = 3

Great work here, everyone!

Here's what I think of as the major takeaway from this problem (and it's something everyone who solved it clearly already knows ) --

I've seen students get tripped up by this and many other problems (in all math sub-topics) because they're trying to get to the answer DIRECTLY. On this problem, obviously if you try to get directly to s(101), it's pretty impossible to just generate an equation for it that'll take you all the way there. Instead, you need to go to the basics of the problem -- start small, even though you ultimately won't have to report what the first or second or third values of the sequence are -- and lo and behold, you notice the pattern that lets you extrapolate. Similarly, on many geometry problems, students get stumped trying to move directly to what the question is asking for, but wouldn't get stumped if they (e.g.) backed up and found the radius and then moved forward from there.

The analogy I always think of here is that you're playing catch with your friend and you keep trying to throw the ball across the yard to him and it KEEPS getting deflected by this giant tree branch in the way. You guys can throw that ball back and forth as many times as you want, trying to get it directly to each other (after all, that's your ultimate objective, right?) and *repeatedly* failing... OR you can pause, walk over to the giant tree branch, break it off and toss it aside, and THEN resume your initial position and throw. The clear, easy shot you'll have gained yourself will be well worth the pause to break off the tree branch!