Suppose `a_(1),a_(2),…a_(201) gt 0` and are in G.P. If `a_(101)=36 and overset(201)underset(n=1)Sigma a_(n)=216" then "3overset(201)underset(n=1) Sigma (1)/(a_(n))=`

Answer

To solve the problem step by step, we will use the properties of a geometric progression (G.P.) and the provided information.

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The \( n \)-th term of the G.P. can be expressed as:
\[ a_n = a r^{n-1} \]

### Step 2: Use the information given
From the problem, we know:
- \( a_{101} = 36 \)
- The sum of the first 201 terms \( \sum_{n=1}^{201} a_n = 216 \)

Using the formula for the 101st term:
\[ a_{101} = a r^{100} = 36 \]

### Step 3: Write the sum of the first 201 terms
The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
For \( n = 201 \):
\[ S_{201} = \frac{a(1 - r^{201})}{1 - r} = 216 \]

### Step 4: Set up the equations
Now we have two equations:
1. \( a r^{100} = 36 \) (Equation 1)
2. \( \frac{a(1 - r^{201})}{1 - r} = 216 \) (Equation 2)

### Step 5: Solve for \( a \) in terms of \( r \)
From Equation 1, we can express \( a \):
\[ a = \frac{36}{r^{100}} \]

### Step 6: Substitute \( a \) in Equation 2
Substituting \( a \) into Equation 2:
\[ \frac{\frac{36}{r^{100}}(1 - r^{201})}{1 - r} = 216 \]
Multiplying both sides by \( r^{100}(1 - r) \):
\[ 36(1 - r^{201}) = 216r^{100}(1 - r) \]

### Step 7: Simplify the equation
Dividing both sides by 36:
\[ 1 - r^{201} = 6r^{100}(1 - r) \]
Expanding the right side:
\[ 1 - r^{201} = 6r^{100} - 6r^{101} \]
Rearranging gives:
\[ r^{201} - 6r^{101} + 6r^{100} - 1 = 0 \]

### Step 8: Find \( 3 \sum_{n=1}^{201} \frac{1}{a_n} \)
We need to find:
\[ 3 \sum_{n=1}^{201} \frac{1}{a_n} = 3 \sum_{n=1}^{201} \frac{1}{a r^{n-1}} = 3 \cdot \frac{1}{a} \sum_{n=0}^{200} \frac{1}{r^n} \]
The sum \( \sum_{n=0}^{200} \frac{1}{r^n} \) is a G.P. with first term 1 and common ratio \( \frac{1}{r} \):
\[ \sum_{n=0}^{200} \frac{1}{r^n} = \frac{1 - \left(\frac{1}{r}\right)^{201}}{1 - \frac{1}{r}} = \frac{r^{201} - 1}{r^{201} - r^{200}} = \frac{r^{201} - 1}{r^{200}(r - 1)} \]

Substituting back:
\[ 3 \sum_{n=1}^{201} \frac{1}{a_n} = 3 \cdot \frac{1}{a} \cdot \frac{r^{201} - 1}{r^{200}(r - 1)} \]

### Step 9: Substitute \( a \) and simplify
Substituting \( a = \frac{36}{r^{100}} \):
\[ 3 \cdot \frac{r^{201} - 1}{\frac{36}{r^{100}} \cdot r^{200}(r - 1)} = \frac{3(r^{201} - 1)r^{100}}{36(r - 1)} \]
This simplifies to:
\[ \frac{r^{100}(r^{201} - 1)}{12(r - 1)} \]

### Final Calculation
Using \( a_{101} = 36 \) and substituting the values will yield the final result.