If `a_(1),a_(2),a_(3), . . .,a_(n)` are non-zero real numbers such that
`(a_(1)^(2)+a_(2)^(2)+ . .. +a_(n-1).^(2))(a_(2)^(2)+a_(3)^(2)+ . . .+a_(n)^(2))le(a_(1)a_(2)+a_(2)a_(3)+ . . . +a_(n-1)a_(n))^(2)" then", a_(1),a_(2), . . . .a_(n)` are in

Options

A. H.P.
B. G.P
C. A.P.
D. none of these

Explanation

To solve the problem, we need to analyze the given inequality and determine if the sequence \( a_1, a_2, \ldots, a_n \) is in Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP).

### Step-by-Step Solution:

1. **Understand the Given Condition**:
The condition given is:
\[ (a_1^2 + a_2^2 + \ldots + a_{n-1}^2)(a_2^2 + a_3^2 + \ldots + a_n^2) \leq (a_1 a_2 + a_2 a_3 + \ldots + a_{n-1} a_n)^2 \]
This is a form of the Cauchy-Schwarz inequality.

2. **Test with \( n = 3 \)**:
Let's take \( n = 3 \) and analyze the sequences for AP, GP, and HP.

- **For Arithmetic Progression (AP)**:
Let \( a_1 = 1, a_2 = 2, a_3 = 3 \).
- Calculate \( LHS \):
\[ LHS = (1^2 + 2^2)(2^2 + 3^2) = (1 + 4)(4 + 9) = 5 \times 13 = 65 \]
- Calculate \( RHS \):
\[ RHS = (1 \cdot 2 + 2 \cdot 3)^2 = (2 + 6)^2 = 8^2 = 64 \]
- Since \( 65 > 64 \), the condition is not satisfied for AP.

- **For Geometric Progression (GP)**:
Let \( a_1 = 1, a_2 = 2, a_3 = 4 \).
- Calculate \( LHS \):
\[ LHS = (1^2 + 2^2)(2^2 + 4^2) = (1 + 4)(4 + 16) = 5 \times 20 = 100 \]
- Calculate \( RHS \):
\[ RHS = (1 \cdot 2 + 2 \cdot 4)^2 = (2 + 8)^2 = 10^2 = 100 \]
- Since \( 100 = 100 \), the condition is satisfied for GP.

- **For Harmonic Progression (HP)**:
Let \( a_1 = 1, a_2 = \frac{1}{2}, a_3 = \frac{1}{3} \).
- Calculate \( LHS \):
\[ LHS = \left(1^2 + \left(\frac{1}{2}\right)^2\right)\left(\left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2\right) = \left(1 + \frac{1}{4}\right)\left(\frac{1}{4} + \frac{1}{9}\right) \]
\[ = \frac{5}{4} \times \left(\frac{9 + 4}{36}\right) = \frac{5}{4} \times \frac{13}{36} = \frac{65}{144} \]
- Calculate \( RHS \):
\[ RHS = \left(1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{3}\right)^2 = \left(\frac{1}{2} + \frac{1}{6}\right)^2 = \left(\frac{3 + 1}{6}\right)^2 = \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]
\[ = \frac{4 \cdot 16}{36} = \frac{64}{144} \]
- Since \( \frac{65}{144} > \frac{64}{144} \), the condition is not satisfied for HP.

3. **Conclusion**:
From the tests, we find that the only progression that satisfies the condition is Geometric Progression (GP).

### Final Answer:
The numbers \( a_1, a_2, \ldots, a_n \) are in **Geometric Progression (GP)**.