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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