If `(a+bk)/(a-bk) = (b+ck)/(b-ck) = (c-dk)/(c-dk)` where `k != 0 , a, b , c d gt 0 ` and none of them equals 1 , then

Options

A. a,b,c,d are in A.P
B. `log_(a)e , log_(b)e , log_(c)e,log_(d)e` are in H.P
C. `b/a = c/b = d/c = 1/k`
D. d,c,b,a are in G.P

Answer

To solve the given problem, we start with the equation:

\[ \frac{a + bk}{a - bk} = \frac{b + ck}{b - ck} = \frac{c + dk}{c - dk} \]

Let's denote this common ratio as \( x \). Therefore, we can write:

1. \(\frac{a + bk}{a - bk} = x\)
2. \(\frac{b + ck}{b - ck} = x\)
3. \(\frac{c + dk}{c - dk} = x\)

### Step 1: Cross-multiply the first equation

From the first equation:

\[ a + bk = x(a - bk) \]

Expanding this gives:

\[ a + bk = ax - bkx \]

Rearranging terms:

\[ a - ax + bk + bkx = 0 \]

Factoring out \( a \) and \( k \):

\[ a(1 - x) + bk(1 + x) = 0 \]

### Step 2: Solve for \( x \)

From the equation above, we can express \( x \):

\[ a(1 - x) = -bk(1 + x) \]

This leads to:

\[ a - ax = -bk - bkx \]

Rearranging gives:

\[ ax + bkx = a + bk \]

Factoring out \( x \):

\[ x(a + bk) = a + bk \]

Thus, we find:

\[ x = 1 \quad \text{(not possible since } a, b, c, d \neq 1) \]

### Step 3: Repeat for the second equation

Now consider the second equation:

\[ b + ck = x(b - ck) \]

Expanding this gives:

\[ b + ck = bx - ckx \]

Rearranging leads to:

\[ b - bx + ck + ckx = 0 \]

Factoring out \( b \) and \( k \):

\[ b(1 - x) + ck(1 + x) = 0 \]

### Step 4: Solve for \( x \) in terms of \( b \) and \( c \)

Similar to before, we can express \( x \):

\[ b(1 - x) = -ck(1 + x) \]

This leads to:

\[ b - bx = -ck - ckx \]

Rearranging gives:

\[ bx + ckx = b + ck \]

Factoring out \( x \):

\[ x(b + ck) = b + ck \]

Thus, we find:

\[ x = 1 \quad \text{(again not possible)} \]

### Step 5: Find the relationship between \( a, b, c, d \)

From both equations, we can derive:

\[ \frac{a}{bk} = \frac{b}{ck} = \frac{c}{dk} \]

Let’s denote this common ratio as \( \lambda \):

\[ \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \lambda \]

This implies:

\[ b^2 = ac \quad \text{and} \quad c^2 = bd \]

### Step 6: Conclusion

From the above relationships, we conclude that:

- \( a, b, c \) are in geometric progression (GP).
- \( b, c, d \) are also in GP.

Thus, the final conclusion is:

- \( a, b, c, d \) are in GP.