j_shreyans wrote:If an ≠ 0 and n is a positive integer, is n odd?

(1) a^n + a^n+1 < 0

(2) a is an integer.

Statement 1: a^n + a^(n+1) < 0
This inequality holds true only if a<0.

Test one case that also satisfies statement 2 and one case that does not.
It's possible that a=-2 and n=2, since (-2)² + (-2)³ = -4.
It's possible that a=-1/2 and n=1, since (-1/2)¹ + (-1/2)² = -1/4.
Since n is EVEN in the first case but ODD in the second case, INSUFFICIENT.

Statement 2: a is an integer
No information about n.
INSUFFICIENT.

Statements combined:
a must be a negative integer.
If n is odd, then a^n + a^(n+1) = nonnegative.
To illustrate:
If a=-1 and n=1, then a^n + a^(n+1) = (-1)¹ + (-1)² = 0.
If a=-2 and n=3, then a^n + a^(n+1) = (-2)³ + (-2)� = 8.
If a=-3 and n=3, then a^n + a^(n+1) = (-3)³ + (-3)� = 54.
Thus, to satisfy the constraint that a^n + a^(n+1) < 0, n CANNOT be odd.
SUFFICIENT.

The correct answer is C.

j_shreyans wrote:If an ≠ 0 and n is a positive integer, is n odd?

(1) a^n + a^n+1 < 0

(2) a is an integer.

OAC

Is my below solution right? ^^

If an ≠ 0 and n is a positive integer, is n odd?

(1) a^n + a^n+1 < 0

(2) a is an integer

Think if the logic this way: This is basically testing even/odd powers of a negative number concept.

1) a^n (1 + a) <0..now can a>0? NO as a^n >0 and (+a)>0. So, a <0. There is a reason why they are asking whether n is odd. We want to find out whether a^n is positive or negative (give that a<0).
now, since we don't know what a is (integer or not), (1+a) can be <0 or >0, giving us no clue whether a^n is positive or negative (or n even or odd, given that a<0).

Statement 2 tells us exactly that a integer,,,,we know a<0 so, (1+a) <0 => (-ve)^n needs to be positive. So, we know n has to be even.