If `a ,b, c ,d` are in G.P, then `(b-c)^2+(c-a)^2+(d-b)^2` is equal to `

Options

A. `(a-d)^(2)`
B. `(ad)^(2)`
C. `(a+d)^(2)`
D. `(a//d)^(2)`

Explanation

To solve the problem, we need to show that if \( a, b, c, d \) are in geometric progression (G.P.), then the expression \( (b-c)^2 + (c-a)^2 + (d-b)^2 \) simplifies to \( (a - d)^2 \).

### Step-by-step Solution:

1. **Define the terms in G.P.**:
Since \( a, b, c, d \) are in G.P., we can express them in terms of a common ratio \( r \):
- Let \( b = ar \)
- Let \( c = ar^2 \)
- Let \( d = ar^3 \)

2. **Substitute the values into the expression**:
We need to evaluate:
\[ (b-c)^2 + (c-a)^2 + (d-b)^2 \]
Substituting the values of \( b, c, d \):
\[ (ar - ar^2)^2 + (ar^2 - a)^2 + (ar^3 - ar)^2 \]

3. **Simplify each term**:
- First term:
\[ (ar - ar^2)^2 = (ar(1 - r))^2 = a^2 r^2 (1 - r)^2 \]
- Second term:
\[ (ar^2 - a)^2 = (a(r^2 - 1))^2 = a^2 (r^2 - 1)^2 \]
- Third term:
\[ (ar^3 - ar)^2 = (ar(r^2 - 1))^2 = a^2 r^2 (r^2 - 1)^2 \]

4. **Combine the terms**:
Now we combine all the terms:
\[ a^2 r^2 (1 - r)^2 + a^2 (r^2 - 1)^2 + a^2 r^2 (r^2 - 1)^2 \]
Factor out \( a^2 \):
\[ a^2 \left[ r^2 (1 - r)^2 + (r^2 - 1)^2 + r^2 (r^2 - 1)^2 \right] \]

5. **Simplify the expression inside the brackets**:
- The first term \( r^2 (1 - r)^2 \) expands to \( r^2 (1 - 2r + r^2) = r^2 - 2r^3 + r^4 \).
- The second term \( (r^2 - 1)^2 \) expands to \( r^4 - 2r^2 + 1 \).
- The third term \( r^2 (r^2 - 1)^2 \) expands to \( r^2 (r^4 - 2r^2 + 1) = r^6 - 2r^4 + r^2 \).

Combining these gives:
\[ (r^2 - 2r^3 + r^4) + (r^4 - 2r^2 + 1) + (r^6 - 2r^4 + r^2) \]
Simplifying further:
\[ r^6 - 2r^3 + 0 + 1 = r^6 - 2r^3 + 1 \]

6. **Final expression**:
Thus, we have:
\[ (b-c)^2 + (c-a)^2 + (d-b)^2 = a^2 (1 - r^3)^2 \]
Since \( d = ar^3 \), we can write:
\[ = (a - d)^2 \]

### Conclusion:
Thus, we conclude that:
\[ (b-c)^2 + (c-a)^2 + (d-b)^2 = (a - d)^2 \]