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.
 A: HINT: 
Since  it is in G.P. you can take $a_i = k.a^i$. Simplify $\sum \frac{1}{a_i}$ by taking a common denominator. Similarly simplify P. Rest will become obvious.
A: 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! :)
