Let 'a' be the first term and 'r' be the common ratio.

To solve this problem, let's assume that the first term of the infinite geometric progression (GP) is 'a' and the common ratio is 'r'.

Sum of the first two terms:
The first term is 'a' and the second term is 'ar' (since it's a GP). Given that the sum of the first two terms is 1, we have the equation:
a + ar = 1

Every term is twice the sum of the successive terms:
Let's consider the third term of the GP. It is 'ar^2'. According to the given condition, every term is twice the sum of the successive terms. Therefore, we have:
2(ar + ar^2) = ar^2

Simplifying the equation:
2ar + 2ar^2 = ar^2
2ar^2 - ar^2 - 2ar = 0
ar^2 - 2ar - ar^2 = 0
-2ar = 0

Since 'r' cannot be zero (as it's a common ratio), we can divide both sides of the equation by 'r':
-2a = 0

Solving for 'a':
a = 0/-2 = 0

Now, substituting the value of 'a' back into the equation for the sum of the first two terms:
0 + 0r = 1
0r = 1

Since any number multiplied by zero is zero, there is no value of 'r' that satisfies this equation. Therefore, the assumption that 'a' is zero is incorrect.

So, the correct answer is the option that does not assume 'a' to be zero.

Let's check all the options one by one:

a) 1/3:
If 'a' is 1/3, then the second term is 1/3 * r. The sum of the first two terms is:
1/3 + 1/3 * r = 1
r + 1 = 3r
2r = 1
r = 1/2

Now, let's check if every term is twice the sum of the successive terms:
First term = 1/3
Second term = 1/3 * 1/2 = 1/6
Third term = 1/3 * (1/2)^2 = 1/12

Is every term twice the sum of the successive terms?
1/3 = 2 * (1/6)
1/6 = 2 * (1/12)

Both equations satisfy the condition, so option 'a' is a valid solution.

b) 2/3:
If 'a' is 2/3, then the second term is 2/3 * r. The sum of the first two terms is:
2/3 + 2/3 * r = 1
2r + 3 = 9r
7r = 3
r = 3/7

Now, let's check if every term is twice the sum of the successive terms:
First term = 2/3
Second term = 2/3 * 3/7 = 2/7
Third term = 2/3 * (3/7)^2 = 18/147

Is every term twice the sum of the successive terms?
2/3 = 2