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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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.

\(k\) 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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(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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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.

Couldnt k itself be negative?

Thank you! I think i've finally got it.