Final answer:
The 31st smallest term and the 16th largest term are both zero because the sum of the first 31 terms is zero in an arithmetic progression. The sum of the largest and smallest terms is zero, not positive. Thus, statements I and II are correct, making statement III incorrect.
Step-by-step explanation:
Let's analyze the given arithmetic progression (AP). In an AP, the sum of equal numbers of terms equidistant from the beginning and end of the sequence is the same. Therefore, if the sum of the first 31 terms is zero, the sum of these 31 terms and the remaining 15 terms must also be zero, since the whole sequence would sum to zero. It means that the sum of the last 15 terms is also zero.
Now, recall that the average value of an AP is the middle term when the number of terms is odd, or it is the average of the two middle terms when the number of terms is even.
Since we have an odd number of terms (31), the 16th term is the middle term of the first 31 terms.
The fact that the sum of the first 31 terms is zero implies that the 16th term must be zero because it is equal to the average value of the sequence, which is the sum divided by the number of terms (0/31=0).
But since it's also the 31st largest term, statement I and statement II are correct.
Regarding statement III, it suggests that the sum of the largest and smallest terms is positive, which cannot be concluded from the given information.
Since the sum of the entire sequence is zero, the average of the first and last term (which is equal to the sum of the largest and smallest term divided by 2) will be zero as well.
The sum of the largest and smallest terms is therefore zero, not positive. Thus, statement III is incorrect.