Suppose four distinct positive numbers `a_(1),a_(2),a_(3),a_(4)` are in G.P. Let `b_(1)=a_(1)+,a_(b)=b_(1)+a_(2),b_(3)=b_(2)+a_(3)andb_(4)=b_(3)+a_(4)`.
Statement -1 : The numbers `b_(1),b_(2),b_(3),b_(4)` are neither in A.P. nor in G.P.
Statement -2: The numbers `b_(1),b_(2),b_(3),b_(4)` are in H.P.

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1
C. Statement 1 is true, Statement 2 is false
D. Statement 1 is false, Statement 2 is true

Explanation

To solve the problem, we need to analyze the sequences defined by the numbers \(a_1, a_2, a_3, a_4\) in geometric progression (G.P.) and the numbers \(b_1, b_2, b_3, b_4\) defined in terms of \(a_i\).

### Step-by-Step Solution:

1. **Define the G.P.**:
Let the four distinct positive numbers in G.P. be defined as:
\[ a_1 = x, \quad a_2 = xr, \quad a_3 = xr^2, \quad a_4 = xr^3 \]
where \(x\) is a positive number and \(r\) is the common ratio (with \(r > 0\) and \(r \neq 1\)).

2. **Define the \(b_i\) values**:
We will calculate \(b_1, b_2, b_3, b_4\):
- \(b_1 = a_1 = x\)
- \(b_2 = b_1 + a_2 = x + xr = x(1 + r)\)
- \(b_3 = b_2 + a_3 = x(1 + r) + xr^2 = x(1 + r + r^2)\)
- \(b_4 = b_3 + a_4 = x(1 + r + r^2) + xr^3 = x(1 + r + r^2 + r^3)\)

3. **Express \(b_i\) in terms of \(x\)**:
We can summarize the \(b_i\) values as:
\[ b_1 = x, \quad b_2 = x(1 + r), \quad b_3 = x(1 + r + r^2), \quad b_4 = x(1 + r + r^2 + r^3) \]

4. **Check for Arithmetic Progression (A.P.)**:
To check if \(b_1, b_2, b_3, b_4\) are in A.P., we need to see if:
\[ b_2 - b_1 = b_3 - b_2 \quad \text{and} \quad b_3 - b_2 = b_4 - b_3 \]
- Calculate \(b_2 - b_1\):
\[ b_2 - b_1 = x(1 + r) - x = xr \]
- Calculate \(b_3 - b_2\):
\[ b_3 - b_2 = x(1 + r + r^2) - x(1 + r) = xr^2 \]
- Since \(xr \neq xr^2\) (as \(r \neq 1\)), \(b_1, b_2, b_3\) are not in A.P.

5. **Check for Geometric Progression (G.P.)**:
To check if \(b_1, b_2, b_3, b_4\) are in G.P., we need to see if:
\[ \frac{b_2}{b_1} = \frac{b_3}{b_2} = \frac{b_4}{b_3} \]
- Calculate \(\frac{b_2}{b_1}\):
\[ \frac{b_2}{b_1} = \frac{x(1 + r)}{x} = 1 + r \]
- Calculate \(\frac{b_3}{b_2}\):
\[ \frac{b_3}{b_2} = \frac{x(1 + r + r^2)}{x(1 + r)} = \frac{1 + r + r^2}{1 + r} \]
- Since \(1 + r \neq \frac{1 + r + r^2}{1 + r}\) (as \(r \neq 1\)), \(b_1, b_2, b_3\) are not in G.P.

6. **Check for Harmonic Progression (H.P.)**:
The numbers \(b_1, b_2, b_3, b_4\) can only be in H.P. if their reciprocals are in A.P. Since we have already established that they are not in A.P. or G.P., they cannot be in H.P. either.

### Conclusion:
- **Statement 1**: True - The numbers \(b_1, b_2, b_3, b_4\) are neither in A.P. nor in G.P.
- **Statement 2**: False - The numbers \(b_1, b_2, b_3, b_4\) are not in H.P.