Review question

Ref: R9468

## 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

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 $a=d=-1$.