The first term of an infinite G..P is 1 and any term is equal to the sum of all the succeeding terms. Find the series.

Options

A. `1, 1/2, 1/8, 1/16 ,…..oo`
B. `1, 1/8, 1/6, 1/32,…….oo`
C. `1, 1/2, 1/4,1/8,…..oo`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the terms of the geometric progression (G.P.)
Let the first term of the infinite G.P. be \( a = 1 \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
- First term: \( a = 1 \)
- Second term: \( ar = r \)
- Third term: \( ar^2 = r^2 \)
- Fourth term: \( ar^3 = r^3 \)
- And so on...

### Step 2: Write the formula for the sum of the infinite G.P.
The sum \( S \) of an infinite G.P. is given by the formula:
\[ S = \frac{a}{1 - r} \]
Substituting \( a = 1 \):
\[ S = \frac{1}{1 - r} \]

### Step 3: Set up the equation based on the problem statement
According to the problem, any term in the G.P. is equal to the sum of all the succeeding terms. Let's take the first term \( a = 1 \) as an example. The succeeding terms are \( r, r^2, r^3, \ldots \). Therefore, we can write:
\[ 1 = S_{\text{succeeding}} = r + r^2 + r^3 + \ldots \]

### Step 4: Express the sum of the succeeding terms
The sum of the succeeding terms can also be expressed as an infinite G.P.:
\[ S_{\text{succeeding}} = \frac{r}{1 - r} \]

### Step 5: Set the two expressions equal to each other
Now we equate the two expressions:
\[ 1 = \frac{r}{1 - r} \]

### Step 6: Solve for \( r \)
Cross-multiplying gives:
\[ 1 - r = r \]
Rearranging this leads to:
\[ 1 = 2r \implies r = \frac{1}{2} \]

### Step 7: Write the series
Now that we have the common ratio \( r = \frac{1}{2} \), we can write the terms of the G.P.:
- First term: \( 1 \)
- Second term: \( \frac{1}{2} \)
- Third term: \( \frac{1}{4} \)
- Fourth term: \( \frac{1}{8} \)
- And so on...

Thus, the series is:
\[ 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \]

### Final Answer
The series is \( 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \)

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