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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