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

Answer

To solve the problem step by step, let's follow the reasoning laid out in the video transcript.

### Step 1: Define the terms of the G.P.
We know the first term \( a_1 = 1 \). Since it is a geometric progression (G.P.), we can express the subsequent terms as:
- Second term: \( a_2 = ar = 1 \cdot r = r \)
- Third term: \( a_3 = ar^2 = 1 \cdot r^2 = r^2 \)
- Fourth term: \( a_4 = ar^3 = 1 \cdot r^3 = r^3 \)
- And so on...

### Step 2: Set up the equation based on the problem statement
According to the problem, any term is equal to the sum of all the succeeding terms. Let's take the first term \( a_1 \):
\[ a_1 = a_2 + a_3 + a_4 + \ldots \]
This can be expressed as:
\[ 1 = r + r^2 + r^3 + \ldots \]

### Step 3: Use the formula for the sum of an infinite G.P.
The sum \( S \) of an infinite G.P. can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. In our case, the first term \( a = r \) and the common ratio is also \( r \):
\[ S = \frac{r}{1 - r} \]

### Step 4: Set the equation
From the previous steps, we have:
\[ 1 = \frac{r}{1 - r} \]

### Step 5: Solve for \( r \)
Cross-multiplying gives:
\[ 1 - r = r \]
Rearranging this, we get:
\[ 1 = 2r \]
Thus,
\[ r = \frac{1}{2} \]

### Step 6: Write the series
Now that we have \( r = \frac{1}{2} \), we can write the series:
- First term: \( a_1 = 1 \)
- Second term: \( a_2 = r = \frac{1}{2} \)
- Third term: \( a_3 = r^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
- Fourth term: \( a_4 = r^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \)
- And so on...

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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