<s>[INST] finding value of $\sum^{10}_{i=0}\frac{1}{a_{i}}$ Let $a_{0},a_{1},a_{2},a_{3},\cdots \cdots a_{n}$ be a sequence of numbers satisying $(3-a_{n+1})(6+a_{n}) = 18$ and $a_{0} = 3.$ then find $\displaystyle \sum^{10}_{i=0}\frac{1}{a_{i}}$
from $18-6a_{n+1}+3a_{n}-a_{n}a_{n+1} = 18.$ So $6a_{n+1}-3a_{n}+a_{n}a_{n+1} =0$
want be able to go further, could some help me, thanks
 [/INST] From the condition we have $$-6a_{n+1}+3a_n-a_na_{n+1}=0$$ or
$$-\frac{6}{a_n}+\frac{3}{a_{n+1}}-1=0$$ or
$$-6\left(\frac{1}{a_0}+\frac{1}{a_1}+...+\frac{1}{a_{10}}\right)+3\left(\frac{1}{a_1}+...+\frac{1}{a_{11}}\right)-11=0$$
and for the ending it remains to find a value of $a_{11}$.
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