Let `a_(1),a_(2),a_(3),"......"a_(10)` are in GP with `a_(51)=25` and `sum_(i=1)^(101)a_(i)=125 " than the value of " sum_(i=1)^(101)((1)/(a_(i)))` equals.

Options

A. 5
B. `(1)/(5)`
C. `(1)/(25)`
D. `(1)/(125)`

Explanation

To solve the problem step by step, let's denote the first term of the geometric progression (GP) as \( a_1 \) and the common ratio as \( r \).

### Step 1: Write the sum of the first 101 terms of the GP
The sum of the first \( n \) terms of a GP is given by the formula:

\[ S_n = a_1 \frac{1 - r^n}{1 - r} \]

For our case, we need to find the sum of the first 101 terms, so we have:

\[ S_{101} = a_1 \frac{1 - r^{101}}{1 - r} \]

We know from the problem that this sum equals 125:

\[ a_1 \frac{1 - r^{101}}{1 - r} = 125 \quad \text{(Equation 1)} \]

### Step 2: Use the information about \( a_{51} \)
The \( n \)-th term of a GP can be expressed as:

\[ a_n = a_1 r^{n-1} \]

Thus, for \( n = 51 \):

\[ a_{51} = a_1 r^{50} = 25 \quad \text{(Equation 2)} \]

### Step 3: Solve for \( a_1 \) in terms of \( r \)
From Equation 2, we can express \( a_1 \):

\[ a_1 = \frac{25}{r^{50}} \quad \text{(Equation 3)} \]

### Step 4: Substitute \( a_1 \) in Equation 1
Now, substitute Equation 3 into Equation 1:

\[ \frac{25}{r^{50}} \frac{1 - r^{101}}{1 - r} = 125 \]

### Step 5: Simplify the equation
Multiply both sides by \( r^{50} \):

\[ 25(1 - r^{101}) = 125(1 - r)r^{50} \]

Dividing both sides by 25 gives:

\[ 1 - r^{101} = 5(1 - r)r^{50} \]

### Step 6: Rearranging the equation
Rearranging the equation, we get:

\[ 1 - r^{101} = 5r^{50} - 5r^{51} \]

### Step 7: Find the sum of the reciprocals
We need to find:

\[ \sum_{i=1}^{101} \frac{1}{a_i} \]

This can be expressed as:

\[ \sum_{i=1}^{101} \frac{1}{a_1 r^{i-1}} = \frac{1}{a_1} \sum_{i=0}^{100} r^{-i} \]

The sum \( \sum_{i=0}^{100} r^{-i} \) is a geometric series:

\[ \sum_{i=0}^{100} r^{-i} = \frac{1 - r^{-101}}{1 - r^{-1}} = \frac{1 - r^{-101}}{1 - \frac{1}{r}} = \frac{r^{101} - 1}{r^{101}(r - 1)} \]

Thus, we have:

\[ \sum_{i=1}^{101} \frac{1}{a_i} = \frac{1}{a_1} \cdot \frac{r^{101} - 1}{r^{101}(r - 1)} \]

### Step 8: Substitute \( a_1 \) and simplify
Substituting \( a_1 = \frac{25}{r^{50}} \):

\[ \sum_{i=1}^{101} \frac{1}{a_i} = \frac{r^{101} - 1}{\frac{25}{r^{50}} \cdot r^{101}(r - 1)} = \frac{(r^{101} - 1)r^{50}}{25r^{101}(r - 1)} \]

### Step 9: Use \( a_{51} = 25 \)
Since \( a_{51} = 25 \), we know \( a_1 r^{50} = 25 \). Thus, we can find the value of \( \sum_{i=1}^{101} \frac{1}{a_i} \):

\[ \sum_{i=1}^{101} \frac{1}{a_i} = \frac{125}{25^2} = \frac{125}{625} = \frac{1}{5} \]

### Final Answer
The value of \( \sum_{i=1}^{101} \frac{1}{a_i} \) equals \( \frac{1}{5} \).

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