Let `a_n` be the `n^(th)` term, of a G.P of postive terms. If `Sigma_(n=1)^(100)a_(2n+1)= 200` and
` Sigma_(n=1)^(100)a_(2n)=100, " then" Sigma_(n=1)^(200)a_(n)`

Options

A. 150
B. 225
C. 300
D. 175

Answer

To solve the problem, we need to find the sum \( \Sigma_{n=1}^{200} a_n \) given the sums \( \Sigma_{n=1}^{100} a_{2n+1} = 200 \) and \( \Sigma_{n=1}^{100} a_{2n} = 100 \).

### Step 1: Express the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
- \( a_1 = a \)
- \( a_2 = ar \)
- \( a_3 = ar^2 \)
- \( a_4 = ar^3 \)
- ...
- \( a_n = ar^{n-1} \)

### Step 2: Rewrite the given sums
The sum \( \Sigma_{n=1}^{100} a_{2n+1} \) can be rewritten as:
\[ \Sigma_{n=1}^{100} a_{2n+1} = a_3 + a_5 + a_7 + \ldots + a_{201} = ar^2 + ar^4 + ar^6 + \ldots + ar^{200} \]
This is a G.P. with first term \( ar^2 \) and common ratio \( r^2 \) for 100 terms.

The sum can be calculated using the formula for the sum of a G.P.:
\[ \Sigma_{n=1}^{100} a_{2n+1} = ar^2 \frac{(r^2)^{100} - 1}{r^2 - 1} = 200 \]

The sum \( \Sigma_{n=1}^{100} a_{2n} \) can be rewritten as:
\[ \Sigma_{n=1}^{100} a_{2n} = a_2 + a_4 + a_6 + \ldots + a_{200} = ar + ar^3 + ar^5 + \ldots + ar^{199} \]
This is also a G.P. with first term \( ar \) and common ratio \( r^2 \) for 100 terms.

The sum can be calculated as:
\[ \Sigma_{n=1}^{100} a_{2n} = ar \frac{(r^2)^{100} - 1}{r^2 - 1} = 100 \]

### Step 3: Set up the equations
Now we have two equations:
1. \( ar^2 \frac{r^{200} - 1}{r^2 - 1} = 200 \) (Equation 1)
2. \( ar \frac{r^{200} - 1}{r^2 - 1} = 100 \) (Equation 2)

### Step 4: Divide the equations
Dividing Equation 1 by Equation 2:
\[ \frac{ar^2}{ar} = \frac{200}{100} \]
This simplifies to:
\[ r = 2 \]

### Step 5: Substitute \( r \) back into one of the equations
Substituting \( r = 2 \) into Equation 2:
\[ a(2) \frac{(2^2)^{100} - 1}{2^2 - 1} = 100 \]
This simplifies to:
\[ 2a \frac{2^{200} - 1}{3} = 100 \]
\[ 2a (2^{200} - 1) = 300 \]
\[ a (2^{200} - 1) = 150 \]

### Step 6: Find \( \Sigma_{n=1}^{200} a_n \)
Now we can find \( \Sigma_{n=1}^{200} a_n \):
\[ \Sigma_{n=1}^{200} a_n = a + ar + ar^2 + \ldots + ar^{199} \]
This is a G.P. with first term \( a \) and common ratio \( r \) for 200 terms:
\[ \Sigma_{n=1}^{200} a_n = a \frac{r^{200} - 1}{r - 1} \]
Substituting \( r = 2 \):
\[ \Sigma_{n=1}^{200} a_n = a \frac{2^{200} - 1}{2 - 1} = a (2^{200} - 1) \]

Using \( a (2^{200} - 1) = 150 \):
\[ \Sigma_{n=1}^{200} a_n = 150 \]

### Final Answer
Thus, the value of \( \Sigma_{n=1}^{200} a_n \) is \( \boxed{150} \).