easiest way to do is put vallues
you will find out that if n is even all terms cancel out each other
onlyt the last term survives which would be positive.
Similarly if n-th term is positive it can happen only if n is odd and it will be sum too.
I tried with n=3, n=4, n=5 etc.
Answer D.
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DS_SEQ
Source: Beat The GMAT — Data Sufficiency |
The problem is made to look awkward and confusing but the solution is pretty straight forward.
If n is odd then the sum of all elements from a(1) to a(n-1) will be 0 (zero). This is because each pair of elements (odd and even) from a(1) to a(n-1) will cancel each other out. The only remaining element i.e. a(n) will be the final sum. Since n is odd, a(n) is positive. Sufficient.
a(n) being positive indicates the same thing since acc. to given data, a(n) is positive only if n is odd. Following the same logic the sum will be equal to a(n) again. Hence sufficient.
Answer is therefore D (each sufficient individually)
If n is odd then the sum of all elements from a(1) to a(n-1) will be 0 (zero). This is because each pair of elements (odd and even) from a(1) to a(n-1) will cancel each other out. The only remaining element i.e. a(n) will be the final sum. Since n is odd, a(n) is positive. Sufficient.
a(n) being positive indicates the same thing since acc. to given data, a(n) is positive only if n is odd. Following the same logic the sum will be equal to a(n) again. Hence sufficient.
Answer is therefore D (each sufficient individually)
