A geometric sequence has its first element as 6, the sum of the first $n$ elements is $\frac{45}{4}$, and the sum of the reciprocals of these elements is $\frac{5}{2}$. Which is this geometric sequence?
Solution. The first term of the geometric sequence is $a_{1}=6$. The ratio of the sequence is not 1, for if it were, the sum of the first $n$ terms would be: $\frac{45}{4}=6 n$, which would imply $n=\frac{45}{24}$, not an integer. According to the known formula: $S_{n}=a_{1} \frac{q^{n}-1}{q-1}$, thus

$$
6 \cdot \frac{q^{n}-1}{q-1}=\frac{45}{4}
$$

Rearranging the equation:

$$
q^{n}-1=\frac{15}{8}(q-1)
$$

It is known that the reciprocals of the terms of a geometric sequence also form a geometric sequence. The ratio of this sequence is $\frac{1}{q}$, so according to the condition,

$$
\frac{5}{2}=\frac{1}{6} \cdot \frac{\left(\frac{1}{q}\right)^{n}-1}{\frac{1}{q}-1}
$$

Perform the indicated operations. After rearrangement, we get

$$
q^{n}-1=15\left(q^{n}-q^{n-1}\right)
$$

Substitute the expression for $\left(q^{n}-1\right)$ from (1) into (2):

$$
\frac{15}{8}(q-1)=15\left(q^{n}-q^{n-1}\right)=15 q^{n-1}(q-1)
$$

Dividing by $15(q-1) \neq 0$: $\frac{1}{8}=q^{n-1}$, hence $q^{n}=\frac{1}{8} q$. Substituting this into (1) gives $\frac{1}{8} q-1=\frac{15}{8}(q-1)$, from which $q=\frac{1}{2}$ and $\left(\frac{1}{2}\right)^{n}=\frac{1}{16}$, so $n=4$.

The elements of the first sequence are: $6,3, \frac{3}{2}, \frac{3}{4}$; their sum is indeed $\frac{45}{4}$.

The elements of the second sequence are: $\frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \frac{4}{3}$ and $\frac{1}{6}+\frac{1}{3}+\frac{2}{3}+\frac{4}{3}=\frac{15}{6}=\frac{5}{2}$.