The first term of an infinite G.P. is 1 and every term is equals to the sum of the successive terms, then its fourth term will be

Options

A. `(1)/(2)`
B. `(1)/(8)`
C. `(1)/(4)`
D. `(1)/(16)`

Answer

To solve the problem step by step, we will follow the given conditions of the infinite geometric progression (G.P.):

1. **Identify the first term and common ratio**:
- The first term \( a = 1 \).
- Let the common ratio be \( r \).

2. **Write the terms of the G.P.**:
- The terms of the G.P. are \( 1, r, r^2, r^3, \ldots \).

3. **Set up the equation based on the problem statement**:
- According to the problem, every term is equal to the sum of the successive terms. Thus, we can write:
\[ r + r^2 + r^3 + \ldots = 1 \]
- This is the sum of an infinite G.P. starting from the second term.

4. **Use the formula for the sum of an infinite G.P.**:
- The sum \( S \) of an infinite G.P. is given by:
\[ S = \frac{a}{1 - r} \]
- Here, the first term \( a = r \) (the second term of the original G.P.) and the common ratio is also \( r \). Therefore, we have:
\[ r + r^2 + r^3 + \ldots = \frac{r}{1 - r} \]
- Setting this equal to 1 gives:
\[ \frac{r}{1 - r} = 1 \]

5. **Solve for \( r \)**:
- Cross-multiplying gives:
\[ r = 1 - r \]
- Rearranging this yields:
\[ 2r = 1 \implies r = \frac{1}{2} \]

6. **Find the fourth term of the G.P.**:
- The \( n \)-th term of a G.P. is given by:
\[ T_n = a \cdot r^{n-1} \]
- For the fourth term (\( n = 4 \)):
\[ T_4 = 1 \cdot \left(\frac{1}{2}\right)^{4-1} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

7. **Conclusion**:
- Therefore, the fourth term of the G.P. is \( \frac{1}{8} \).