[Algebra]
Problem: A geometric sequence has first term 10 and common ratio $\frac{1}{2}$.

An arithmetic sequence has first term 10 and common difference $d$.

The ratio of the 6th term in the geometric sequence to the 4th term in the geometric sequence equals the ratio of the 6th term in the arithmetic sequence to the 4 th term in the arithmetic sequence.

Determine all possible values of $d$.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, 3, 5, 7, 9 are the first four terms of an arithmetic sequence.

A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, $3,6,12$ is a geometric sequence with three terms.)

Solution: The first 6 terms of a geometric sequence with first term 10 and common ratio $\frac{1}{2}$ are $10,5, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \frac{5}{16}$.

Here, the ratio of its 6 th term to its 4 th term is $\frac{5 / 16}{5 / 4}$ which equals $\frac{1}{4}$. (We could have determined this without writing out the sequence, since moving from the 4th term to the 6th involves multiplying by $\frac{1}{2}$ twice.)

The first 6 terms of an arithmetic sequence with first term 10 and common difference $d$ are $10,10+d, 10+2 d, 10+3 d, 10+4 d, 10+5 d$.

Here, the ratio of the 6 th term to the 4 th term is $\frac{10+5 d}{10+3 d}$.

Since these ratios are equal, then $\frac{10+5 d}{10+3 d}=\frac{1}{4}$, which gives $4(10+5 d)=10+3 d$ and so $40+20 d=10+3 d$ or $17 d=-30$ and so $d=-\frac{30}{17}$.

Final answer: $-\frac{30}{17}$