Solution

The first term of an arithmetic sequence is \(a\) and the common difference is \(d\).

The sum of the first \(n\) terms is denoted by \(S_n\).

If \(S_8>3S_6\), what can be deduced about the sign of \(a\) and the sign of \(d\)?

  1. both \(a\) and \(d\) are negative

  2. \(a\) is positive, \(d\) is negative

  3. \(a\) is negative, \(d\) is positive

  4. \(a\) is negative, but the sign of \(d\) cannot be deduced

  5. \(d\) is negative, but the sign of \(a\) cannot be deduced

  6. neither the sign of \(a\) nor the sign of \(d\) can be deduced

[Choose the one correct answer and explain your reasoning.]

The most appropriate formula for \(S_n\) in this context is \(S_n=\frac{1}{2}n(2a+(n-1)d)\). Substituting \(n=6\) and \(n=8\) gives \[\begin{align*} S_6&=3(2a+5d)=6a+15d\\ S_8&=4(2a+7d)=8a+28d. \end{align*}\]

Now we are told that \(S_8>3S_6\), so \(8a+28d>3(6a+15d)\), which gives \[-10a -17d > 0.\] As this is the only information we have, the only thing we can say about the signs of \(a\) and \(d\) is that at least one of them is negative, but we cannot determine which one. For example, we could have \(a=1\) and \(d=-1\), or \(a=-2\) and \(d=1\), or \(a=d=-1\).

So the correct answer is (F).