The sum of first two terms of an infinite G.P. is 1 and every terms is twice the sum of the successive terms. Its first terms is

Options

A. `1//3`
B. `2//3`
C. `3//4`
D. `1//4`

Answer

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

### Step 1: Set up the equation for the sum of the first two terms
The sum of the first two terms of the G.P. is given as:
\[ a + ar = 1 \]
This can be factored as:
\[ a(1 + r) = 1 \quad \text{(1)} \]

### Step 2: Set up the equation for the relationship between terms
According to the problem, every term is twice the sum of the successive terms. The sum of the terms from the second term onward can be expressed as:
\[ ar + ar^2 + ar^3 + \ldots = \frac{ar}{1 - r} \quad \text{(for } |r| < 1\text{)} \]
Thus, the equation becomes:
\[ a = 2 \left( \frac{ar}{1 - r} \right) \quad \text{(2)} \]

### Step 3: Substitute and simplify
From equation (2), we can substitute \( a \):
\[ a = \frac{2ar}{1 - r} \]
Now, we can simplify this:
\[ 1 = \frac{2r}{1 - r} \quad \text{(dividing both sides by } a \text{, assuming } a \neq 0\text{)} \]
Cross-multiplying gives:
\[ 1 - r = 2r \]
This simplifies to:
\[ 1 = 3r \quad \Rightarrow \quad r = \frac{1}{3} \quad \text{(3)} \]

### Step 4: Substitute \( r \) back into equation (1)
Now, we substitute \( r = \frac{1}{3} \) back into equation (1):
\[ a(1 + \frac{1}{3}) = 1 \]
This simplifies to:
\[ a \cdot \frac{4}{3} = 1 \]
Thus, we find:
\[ a = \frac{3}{4} \quad \text{(4)} \]

### Conclusion
The first term of the infinite G.P. is:
\[ \boxed{\frac{3}{4}} \]