If ` a_(1),a_(2),a_(3),a_(4),,……, a_(n-1),a_(n) " are distinct non-zero real numbers such that " (a_(1)^(2) + a_(2)^(2) + a_(3)^(2) + …..+ a_(n-1)^(2))x^2 + 2 (a_(1)a_(2) + a_(2)a_(3) + a_(3)a_(4) + ……+ a_(n-1) a_(n))x + (a_(2)^(2) +a_(3)^(2) + a_(4)^(2) +......+ a_(n)^(2)) le 0 " then " a_(1), a_(2), a_(3) ,....., a_(n-1), a_(n)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. A.G.P.

Answer

To solve the problem, we need to analyze the given quadratic inequality and deduce the relationships among the distinct non-zero real numbers \( a_1, a_2, a_3, \ldots, a_n \).

### Step-by-Step Solution

1. **Identify the Quadratic Form**:
The given expression can be rewritten in the standard quadratic form:
\[ (a_1^2 + a_2^2 + \ldots + a_{n-1}^2)x^2 + 2(a_1a_2 + a_2a_3 + \ldots + a_{n-1}a_n)x + (a_2^2 + a_3^2 + \ldots + a_n^2) \leq 0 \]
Here, we denote:
- \( A = a_1^2 + a_2^2 + \ldots + a_{n-1}^2 \)
- \( B = 2(a_1a_2 + a_2a_3 + \ldots + a_{n-1}a_n) \)
- \( C = a_2^2 + a_3^2 + \ldots + a_n^2 \)

2. **Condition for Non-positivity**:
For the quadratic \( Ax^2 + Bx + C \leq 0 \) to hold for all \( x \), the following conditions must be satisfied:
- \( A > 0 \) (since \( a_i \) are non-zero)
- The discriminant \( D = B^2 - 4AC \) must be less than or equal to zero.

3. **Calculate the Discriminant**:
We calculate the discriminant:
\[ D = B^2 - 4AC \]
Substituting for \( B \) and \( C \):
\[ D = (2(a_1a_2 + a_2a_3 + \ldots + a_{n-1}a_n))^2 - 4(a_1^2 + a_2^2 + \ldots + a_{n-1}^2)(a_2^2 + a_3^2 + \ldots + a_n^2) \]

4. **Set Up the Inequality**:
For the quadratic to be non-positive, we need:
\[ D \leq 0 \]
This implies that:
\[ (a_1a_2 + a_2a_3 + \ldots + a_{n-1}a_n)^2 \leq (a_1^2 + a_2^2 + \ldots + a_{n-1}^2)(a_2^2 + a_3^2 + \ldots + a_n^2) \]

5. **Apply the Cauchy-Schwarz Inequality**:
The above inequality is a direct application of the Cauchy-Schwarz inequality, which states that:
\[ (x_1y_1 + x_2y_2 + \ldots + x_ny_n)^2 \leq (x_1^2 + x_2^2 + \ldots + x_n^2)(y_1^2 + y_2^2 + \ldots + y_n^2) \]
Here, we can set \( x_i = a_i \) and \( y_i = a_{i+1} \).

6. **Conclude the Relationship**:
The equality condition of the Cauchy-Schwarz inequality holds if and only if the sequences are proportional. This leads us to conclude that:
\[ \frac{a_1}{a_2} = \frac{a_2}{a_3} = \frac{a_3}{a_4} = \ldots = \frac{a_{n-1}}{a_n} \]
This implies that \( a_1, a_2, a_3, \ldots, a_n \) are in geometric progression (GP).

### Final Conclusion:
Thus, the distinct non-zero real numbers \( a_1, a_2, a_3, \ldots, a_n \) are in geometric progression (GP).