Sum of the inverse element geometric sequence

Question

I need assistance with the following question:
$a_1,a_2,....,a_{10} $ are terms of geometric sequence, show formula for the sum of:
$\frac{1}{a_1}+\frac{1}{a_2}+....+\frac{1}{a_{10}}$

I tried to, lowest common denominator and got: $\frac{(a_2a_3...a_{10})+(a_1a_3...a_{10})+...+(a_1a_3...a_{9})}{a_1a_2....a_{10}}=$
$\frac{(a_2a_3...a_{10})+(a_1a_3...a_{10})+...+(a_1a_3...a_{9})}{a_1^{10}qq^2q^3....q^9}$.
but the result should be $\frac{(q^{10}-1)}{a_1q^9(q-1)}$

how they got this answer ?

Answer 1

$\dfrac{1}{a_1}+...+\dfrac{1}{a_{10}}=\dfrac{1}{a_1}+\dfrac{1}{qa_1}+\dfrac{1}{q^2a_1}+...+\dfrac{1}{q^{9}a_1}=\dfrac{1}{a_1}\bigg(1+\dfrac{1}{q}+...+\dfrac{1}{q^9}\bigg)$

Answer 2

Hint : Factor out by the first term, and consider the sequence $u_n = 1/(q^n)$. You should find a result in the form $\sum_k u^k = \frac{u^{n+1}-1}{u-1}$.