The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=-a_{k-1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?

We have following sequence:
\(a_1=1\);
\(a_2=-a_1=-1\);
\(a_3=3\);
\(a_4=-a_3=-3\);
\(a_5=5\);
\(a_6=-a_5=-5\);
...

Basically we have a sequence of positive and negative odd integers: 1, -1, 3, -3, 5, -5, 7., -7, 9, -9, ...

Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(-1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(-1)+3+(-3)=0\). Also notice that odd terms are positive and even terms are negative.

(1) \(n\) is odd --> as discussed the sum is positive. Sufficient.
(2) \(a_n\) is positive --> n is odd, so the same as above. Sufficient.

Answer: D.

Hope it's clear.
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(D) for me

Let us describe the few first terms to have a better idea of it works:
o a(1) = 1
o a(2) = -1
o a(3) = 3
o a(4) = -3

So,
o If n is even, then the sum of a(k) terms give 0. We always have couples of opposite number in the sequence.
o If n is odd, then the sum will be equal to n. All other numbers are in couple (negative/positive), giving 0 if we add them.

From 1
n is odd. Bingo, the sum is positive.

SUFF.

From 2
a(n) > 0... Then n must be an odd. Bingo, the sum is positive.

SUFF.

I'll take a shot at explaining ...

If k is odd we know all the values are +ve and are equal to k (e.g. 1,3,5,7...)
If k is even we know all the values are -ve and are equal to the value of the prior term (e.g. a2 = -1,a4=-3... so the values will be, -1,-3,-5,-7 ....)

1, -1, 3, -3, 5, -5 .....

So as you can see at this point we know that for every value of k (when odd) we have a -ve value from when K is even, unless N (total terms) is odd in which case we will have one extra +ve term that will not cancel out. So if we have even number terms we know the result will be 0 (which is not positive). Therefore to get a positive sum we need one extra odd term.

Try it out,

N=5
1, -1, 3, -3, 5 (if add them, everything cancels out except 5, which is positive).

N=6
1, -1, 3, -3, 5, -5 (if add them, everything cancels out, result is not positive).

So, before looking at the statements we are able to rephrase the question to: "Is the number terms in the sequence odd?"

Statement 1: Gives us exactly that, therefore sufficient.
Statement 2: Well, it gives the same thing but instead of saying the number of terms is odd, it says the last term is +ve, which means the same thing as per our sequence above, there sufficient.

Answer D.

I hope this helps and makes sense.

Now this is what i dont understand. They just mention "k" (i guess constant) but k can take any value -1,-3 or +2. SHouldnt the answer be B then because the statement 2 specifically says that an is positive.

\(k\) in \(a_k\) is a subscript, meaning that \(a_k\) is \(k_{th}\) term in the given sequence which starts from \(a_1\), thus k must be some positive integer.

Complete solution is here: the-sequence-a1-a2-a3-an-of-n-integers-is-such-that-76926.html#p1162192

Hope it helps.
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dont go into complex things. Just visualize the sequence

It can be 2 way

1,-1, 3,-3, 5,-5, 7,-7 ending in negative term The sum will be zero in this case

1,-1, 3,-3, 5,-5, 7 ending in positive term The sum will be the last term of sequence

we have asked is the sum positive ? ----------> is the sequence as per 2nd case ? ----------> is the a(n) odd ? or is the a(n) positive ? both the statements answer these questions so both are sufficient
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(1) n is odd
If n is 5, for instance, then a5=5, a4=-3, a3=3, a2=-1, a1=1 (I used each of the equations listed above to get these numbers I just followed the pattern) When added together [5+(-3)+(3)+(-1)+(1)] the result is +ve5
You could also plug-in 7 for n to make sure and you would still get a +ve result-follow the same pattern.
This statement alone is sufficient.

(2) an is +ve/ means the result must be +ve not zero or -ve

If n=5 (you get the same result as in statement 1)
However, if you let n=4/ following the same procedure as above (statement 1) you will see that the result is 0; therefore, one can conclude that n must be an odd number to get a +ve result. n cannot be even. This statement alone is also sufficient.

The answer is D.
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The sequence a1, a2, a3, ... an of n integers is such that ak=k if k is odd, and ak=−a(k-1) if k is even. Is the sum of the terms in the sequence positive?

(1) n is odd.
-->n = a(n)
For example:
a1 = 1
a2 = -1
a3 = 3
--->Sum of a1; a2; a3 = 3--->a positive integer-->The sum of odd terms is positive--->Sufficient

(2) a(n) is positive
--->n is odd--->The total number is odd--->The sum of odd terms is positive--->Sufficient
--->The answer is D

What i dont understand is the assumption that the sequence starts with a positive term i.e. a1 is positive. If i start with a1= -1, a2= 1, a3 = -3 then the sum is negative. If i start with a1 = 1 the sum is positive. And if i take the values to be even, the sum is zero.

From statement 1, we know that n is odd. Therefore the sum is not 0, but it could be positive or negative. INSUFFICIENT.

From statement 2, we know An is positive, meaning there are odd terms in the sequence, with the last term being positive. SUFFICIENT.

So B should be the answer. Please tell me where i'm wrong in this solution.

We are given that \(a_k=k\), so \(a_1=1\) only.
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Couldnt k itself be negative?

k there is an index number, a subscript indicating the position of a number in the sequence. There cannot be a sequence member with a position of -1. Also, we are given the the sequence is a1, a2, a3, ..., an, .... So, it starts with a1.
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Thank you! I think i've finally got it.

Hi Bunuel - i too got confused with the sub-script k. I was struggling to understand what the difference between subscript "k" and "n" is specifically

Just wondering -- if the question had instead the following theorem to determine the numbers in the sequence -- the answer would be the same right ?

Not \(a_k\) = k per the question

but instead \(a_n\) = n

Would the answer change ?

I dont think there is any difference between "k" and "n"

If no difference, why won't the GMAT just give us \(a_n\) = n instead of \(a_k\) = k ?

In the hopes to trick us perhaps ?

I think it's done to confuse people.

However we know that the question asks about \(n\) integers, where \(a_{some integer}\) = \(some\) \(integer\)

We just need to connect the dots here.
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I always thought of 0 as a non-negative integer.

if the sum is zero, can we assume the sum is positive? Thank you

0 is neither positive nor negative. If the sum is 0, the sum therefore is neither positive nor negative.
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Since n represents the number of terms, n has to be a positive integer. Likewise, k also represent the position of a term in the sequence, therefore, k also represents a positive integer. This is a hidden clue given by the question.

Therefore, if k is odd, we know that \(a_k\) will be a positive odd integer; and if k is even, \(a_k\) will be a negative odd integer.

Applying the rule given,
\(a_1\) = 1, \(a_3\) = 3, \(a_5\) = 5 and so on
\(a_2\) = -\(a_1\) = -1; \(a_4\) = -\(a_3\) = -3; \(a_6\) = -\(a_5\) = -5 and so on

From statement I alone, n is odd. This means the total number of terms are odd and so, the number of odd numbered terms will be higher than the number of even valued terms.
Therefore, the sum of the positives will outweigh the sum of the negatives and hence the sum of the terms of the sequence will be positive.

For example, if n = 5, sum of terms of the sequence = 1 – 1 + 3 – 3 + 5 = 5
Statement I alone is sufficient. Answer options B, C and E can be eliminated.

From statement II alone, \(a_n\) is positive. This means that \(a_n\) is an odd-numbered term and hence n is positive.

From statement I alone, we know that if n is odd, the sum of the terms of the sequence is positive.
Statement II alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
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What if A1 is negative and AN is positive? How do you treat 0 in the series? Would it be odd or even?