Let `a_1,a_2,a_3….., a_(101)` are in G.P with `a_(101) =25 ` and `Sigma_(i=1)^(201) a_i=625` Then the value of `Sigma_(i=1)^(201) 1/a_i` eaquals _______.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the terms of the G.P.
Let the first term of the geometric progression (G.P.) be \( a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
- \( a_1 = a \)
- \( a_2 = ar \)
- \( a_3 = ar^2 \)
- ...
- \( a_{101} = ar^{100} \)

Given that \( a_{101} = 25 \), we have:
\[ ar^{100} = 25 \]

### Step 2: Use the sum of the first 201 terms of the G.P.
The sum of the first \( n \) terms of a G.P. is given by the formula:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
For \( n = 201 \), the sum is:
\[ S_{201} = a \frac{1 - r^{201}}{1 - r} \]
We know from the problem statement that:
\[ S_{201} = 625 \]
Thus, we can write:
\[ a \frac{1 - r^{201}}{1 - r} = 625 \]

### Step 3: Solve for \( a \) and \( r \)
From the equation \( ar^{100} = 25 \), we can express \( a \) in terms of \( r \):
\[ a = \frac{25}{r^{100}} \]
Substituting this into the sum equation:
\[ \frac{25}{r^{100}} \frac{1 - r^{201}}{1 - r} = 625 \]
Multiplying both sides by \( r^{100} \):
\[ 25(1 - r^{201}) = 625 r^{100}(1 - r) \]
Dividing both sides by 25:
\[ 1 - r^{201} = 25 r^{100}(1 - r) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 1 - r^{201} = 25r^{100} - 25r^{101} \]
This can be rearranged to:
\[ r^{201} - 25r^{101} + 25r^{100} - 1 = 0 \]

### Step 5: Find the value of \( \Sigma_{i=1}^{201} \frac{1}{a_i} \)
The sum \( \Sigma_{i=1}^{201} \frac{1}{a_i} \) can be expressed as:
\[ \Sigma_{i=1}^{201} \frac{1}{a_i} = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \ldots + \frac{1}{ar^{200}} \]
This is also a G.P. with first term \( \frac{1}{a} \) and common ratio \( \frac{1}{r} \):
\[ \Sigma_{i=1}^{201} \frac{1}{a_i} = \frac{\frac{1}{a} (1 - \left(\frac{1}{r}\right)^{201})}{1 - \frac{1}{r}} = \frac{1}{a} \cdot \frac{1 - \frac{1}{r^{201}}}{1 - \frac{1}{r}} = \frac{1}{a} \cdot \frac{r^{201} - 1}{r^{201} - r^{200}} \]
Substituting \( a = \frac{25}{r^{100}} \):
\[ \Sigma_{i=1}^{201} \frac{1}{a_i} = \frac{r^{100}}{25} \cdot \frac{r^{201} - 1}{r^{201} - r^{200}} = \frac{r^{100}}{25} \cdot \frac{r^{201} - 1}{r^{200}(r - 1)} = \frac{r^{100}(r^{201} - 1)}{25r^{200}(r - 1)} \]

### Step 6: Substitute \( r^{201} \) and simplify
From earlier, we know \( r^{201} - 25r^{101} + 25r^{100} - 1 = 0 \). We can use this to find \( \Sigma_{i=1}^{201} \frac{1}{a_i} \).

After substituting and simplifying, we find that:
\[ \Sigma_{i=1}^{201} \frac{1}{a_i} = 1 \]

### Final Answer:
Thus, the value of \( \Sigma_{i=1}^{201} \frac{1}{a_i} \) equals **1**.

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