Given that sequence of number `a _(1), a _(2) , a_(3),……, a _(1005)` which satisfy `(a_(1))/( a _(1)+ 1) =(a_(2))/(a _(2) +3) = (a _(3))/(a _(3)+5) = …… = (a _(1005))/(a _(1005)+2009)`
`a_1+a_2+a_3.......a_(1005)=2010`

Options

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

Answer

To solve the problem, we need to analyze the given sequence and the conditions provided. Let's break it down step by step.

### Step 1: Understand the given condition
We are given that:
\[ \frac{a_1}{a_1 + 1} = \frac{a_2}{a_2 + 3} = \frac{a_3}{a_3 + 5} = \ldots = \frac{a_{1005}}{a_{1005} + 2009} \]
Let us denote this common ratio as \( k \). Therefore, we can write:
\[ \frac{a_n}{a_n + (2n - 1)} = k \quad \text{for } n = 1, 2, \ldots, 1005 \]

### Step 2: Express \( a_n \) in terms of \( k \)
From the equation, we can rearrange it to express \( a_n \):
\[ a_n = k(a_n + (2n - 1)) \]
This simplifies to:
\[ a_n = ka_n + k(2n - 1) \]
Rearranging gives:
\[ a_n - ka_n = k(2n - 1) \]
\[ a_n(1 - k) = k(2n - 1) \]
Thus,
\[ a_n = \frac{k(2n - 1)}{1 - k} \]

### Step 3: Find the sum of the sequence
We need to find the sum of the first 1005 terms:
\[ a_1 + a_2 + a_3 + \ldots + a_{1005} = \sum_{n=1}^{1005} a_n = \sum_{n=1}^{1005} \frac{k(2n - 1)}{1 - k} \]
This can be factored out:
\[ = \frac{k}{1 - k} \sum_{n=1}^{1005} (2n - 1) \]
The sum of the first \( N \) odd numbers is \( N^2 \), so:
\[ \sum_{n=1}^{1005} (2n - 1) = 1005^2 \]
Thus, we have:
\[ a_1 + a_2 + a_3 + \ldots + a_{1005} = \frac{k}{1 - k} \cdot 1005^2 \]

### Step 4: Set the sum equal to 2010
According to the problem, we know:
\[ \frac{k}{1 - k} \cdot 1005^2 = 2010 \]
This leads to:
\[ k \cdot 1005^2 = 2010(1 - k) \]
Expanding this gives:
\[ k \cdot 1005^2 = 2010 - 2010k \]
Rearranging terms:
\[ k(1005^2 + 2010) = 2010 \]
Thus:
\[ k = \frac{2010}{1005^2 + 2010} \]

### Step 5: Conclusion
Now we have expressed \( k \) in terms of known quantities. The sequence \( a_n \) can be expressed in terms of \( k \) and the terms are based on the arithmetic progression derived from the sequence structure.