To solve the problem, we start with the given equation:
\[
\frac{a + bk}{a - bk} = \frac{b + ck}{b - ck} = \frac{c + dk}{c - dk}
\]
Let this common value be \( \lambda \). Thus, we can write:
\[
\frac{a + bk}{a - bk} = \lambda
\]
Cross-multiplying gives:
\[
a + bk = \lambda (a - bk)
\]
Expanding this, we have:
\[
a + bk = \lambda a - \lambda bk
\]
Rearranging the terms leads to:
\[
a - \lambda a + bk + \lambda bk = 0
\]
Factoring out \( a \) and \( k \):
\[
a(1 - \lambda) + bk(1 + \lambda) = 0
\]
From this, we can express \( a \) in terms of \( b \) and \( k \):
\[
a(1 - \lambda) = -bk(1 + \lambda)
\]
Now, let’s consider the second equation:
\[
\frac{b + ck}{b - ck} = \lambda
\]
Following the same steps as above, we cross-multiply:
\[
b + ck = \lambda (b - ck)
\]
Expanding gives:
\[
b + ck = \lambda b - \lambda ck
\]
Rearranging leads to:
\[
b - \lambda b + ck + \lambda ck = 0
\]
Factoring out \( b \) and \( k \):
\[
b(1 - \lambda) + ck(1 + \lambda) = 0
\]
Now we have two equations:
1. \( a(1 - \lambda) + bk(1 + \lambda) = 0 \)
2. \( b(1 - \lambda) + ck(1 + \lambda) = 0 \)
Next, we consider the third equation:
\[
\frac{c + dk}{c - dk} = \lambda
\]
Cross-multiplying gives:
\[
c + dk = \lambda (c - dk)
\]
Expanding gives:
\[
c + dk = \lambda c - \lambda dk
\]
Rearranging leads to:
\[
c - \lambda c + dk + \lambda dk = 0
\]
Factoring out \( c \) and \( k \):
\[
c(1 - \lambda) + dk(1 + \lambda) = 0
\]
Now we have three equations:
1. \( a(1 - \lambda) + bk(1 + \lambda) = 0 \)
2. \( b(1 - \lambda) + ck(1 + \lambda) = 0 \)
3. \( c(1 - \lambda) + dk(1 + \lambda) = 0 \)
From these equations, we can express \( a, b, c, d \) in terms of each other.
Dividing the first equation by the second gives:
\[
\frac{a(1 - \lambda) + bk(1 + \lambda)}{b(1 - \lambda) + ck(1 + \lambda)} = 0
\]
This leads to:
\[
\frac{a}{b} = \frac{ck(1 + \lambda)}{bk(1 + \lambda)}
\]
Thus, we can conclude:
\[
\frac{a}{b} = \frac{c}{d}
\]
This implies that \( a, b, c, d \) are in geometric progression (GP).
Next, we can express the relationships:
\[
b^2 = ac \quad \text{and} \quad c^2 = bd
\]
Since \( a, b, c, d \) are in GP, we can also state that the logarithms of these values are in arithmetic progression (AP):
\[
\log a, \log b, \log c, \log d \text{ are in AP}
\]
From the properties of AP, we can conclude that the reciprocals of these values are in harmonic progression (HP):
\[
\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \text{ are in HP}
\]
### Summary of Results:
1. \( a, b, c, d \) are in GP.
2. \( \log a, \log b, \log c, \log d \) are in AP.
3. The reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \) are in HP.
