linear algebra - A problem regarding geometric progressions

Hello my homework included this problem and I really need a hint how to solve it.

It says that the numbers $a_1,a_2 \ldots a_n$ form a geometric progression. Knowing $S=a_1+a_2+\ldots+a_n$ and
$P=a_1 \cdot a_2 \cdot a_3\ldots \cdot a_n$, find $S_1= \frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}$.

I somehow need to find a combination of $S$ and $P$ to form $S_1$ I guess.

Answer

Since $a_1,a_2,a_3,\cdots,a_n$ form a geometric progression then
$$a_k=a_1 r^{k-1}\quad;\;\text{for}\;k=1,2,\cdots,n.$$
Therefore
$$\begin{align} S&=a_1+a_2+a_3+\cdots+a_n\\ &=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}\\ &=a_1\left(1+r+r^2+\cdots+r^{n-1}\right)\\ &=a_1\left(\frac{1-r^n}{1-r}\right), \end{align}$$

$$\begin{align} P&=a_1\cdot a_2\cdot a_3\cdots a_n\\ &=a_1 \cdot a_1r \cdot a_1r^2 \cdots a_1r^{n-1}\\ &=a_1^n r^{0+1+2+\cdots+(n-1)} \\ &=a_1^n r^{\frac n2(n-1)}, \\ \end{align}$$
and

$$\begin{align} S_1&=\frac1a_1+\frac1a_2+\frac1a_3+\cdots+\frac1a_n\\ &=\frac1a_1+\frac1{a_1r}+\frac1{a_1r^2}+\cdots+\frac1{a_1r^{n-1}}\\ &=\frac1a_1\left(1+\frac1r+\frac1{r^2}+\cdots+\frac1{r^{n-1}}\right)\\ &=\frac1a_1\left(\frac{1-\frac1{r^n}}{1-\frac1r}\right)\\ &=\frac{1}{a_1r^{n-1}}\left(\frac{r^n-1}{r-1}\right).\\ \end{align}$$
Now, it should be easy to obtain $S_1$ in term of $S$ and $P$. The rest, I leave to you to handle it. Good luck! :)