Find the G.P. if the sum of its first two terms is 5 and each term is equal to 3 times the sum of its succeeding terms.

Explanation

To solve the problem, we need to find a geometric progression (G.P.) where the sum of the first two terms is 5, and each term is equal to 3 times the sum of its succeeding terms.

Let the first term of the G.P. be \( a \) and the common ratio be \( r \).

### Step 1: Set up the equation for the sum of the first two terms
The first two terms of the G.P. are \( a \) and \( ar \). According to the problem, their sum is given by:
\[ a + ar = 5 \]
Factoring out \( a \), we have:
\[ a(1 + r) = 5 \quad \text{(1)} \]

### Step 2: Set up the equation for the relationship between terms and their succeeding terms
Each term in the G.P. is equal to 3 times the sum of its succeeding terms. The succeeding terms after the first term \( a \) are \( ar, ar^2, ar^3, \ldots \). The sum of the succeeding terms can be expressed as:
\[ ar + ar^2 + ar^3 + \ldots = ar(1 + r + r^2 + \ldots) \]
The sum of an infinite G.P. is given by \( \frac{a}{1 - r} \) if \( |r| < 1 \). Therefore, the sum of the succeeding terms is:
\[ \frac{ar}{1 - r} \]
According to the problem, we have:
\[ a = 3 \times \frac{ar}{1 - r} \quad \text{(2)} \]

### Step 3: Substitute equation (1) into equation (2)
From equation (1), we can express \( a \) in terms of \( r \):
\[ a = \frac{5}{1 + r} \]
Substituting this into equation (2):
\[ \frac{5}{1 + r} = 3 \times \frac{\frac{5}{1 + r} \cdot r}{1 - r} \]
Simplifying the right side:
\[ \frac{5}{1 + r} = \frac{15r}{(1 + r)(1 - r)} \]
Cross-multiplying gives:
\[ 5(1 - r) = 15r \]
Expanding and rearranging:
\[ 5 - 5r = 15r \]
\[ 5 = 20r \]
\[ r = \frac{1}{4} \quad \text{(3)} \]

### Step 4: Substitute \( r \) back to find \( a \)
Now we can substitute \( r = \frac{1}{4} \) back into equation (1) to find \( a \):
\[ a(1 + \frac{1}{4}) = 5 \]
\[ a \cdot \frac{5}{4} = 5 \]
\[ a = 5 \cdot \frac{4}{5} = 4 \quad \text{(4)} \]

### Step 5: Write the G.P.
Now that we have \( a \) and \( r \), we can write the G.P.:
The first term is \( a = 4 \) and the common ratio is \( r = \frac{1}{4} \). Therefore, the G.P. is:
\[ 4, \quad 4 \cdot \frac{1}{4} = 1, \quad 4 \cdot \left(\frac{1}{4}\right)^2 = \frac{1}{4}, \quad 4 \cdot \left(\frac{1}{4}\right)^3 = \frac{1}{16}, \ldots \]
Thus, the G.P. is:
\[ 4, 1, \frac{1}{4}, \frac{1}{16}, \ldots \]

### Final Answer:
The G.P. is \( 4, 1, \frac{1}{4}, \frac{1}{16}, \ldots \)