The 4th term of a G.P.

Question

The 4th term of a G.P. is $500$ and its common ratio is $\dfrac1m$, $m\in\mathbb{N}$. Let $S_n$ denote the sum of the first $n$ terms of this G.P. If $S_6>S_5+1$ and $S_7<S_6+\dfrac12$, then the number of possible values of $m$ is \_\_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

$ar^3=500$ with $r=1/m$. $S_6-S_5=a_6=ar^5>1$; $S_7-S_6=a_7< frac12$. Since $a_6=ar^5=\dfrac{500}{r^{-2}}\cdot\dots$, working through: $a_6>1 \Rightarrow \dfrac{500}{m^2}>1 \Rightarrow m^2<500 \Rightarrow m<22.36$; $a_7< frac12 \Rightarrow \dfrac{500}{m^3}< frac12 \Rightarrow m^3>1000 \Rightarrow m>10$. So $m \in \{11,12,\ldots,22\}$ --- 12 possible values.

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