The first term of a G.P. which is 2 more than the second term and the sum of infinilty is 50. The first term of G.P. is

Options

A. 10
B. `-10`
C. `+-10`
D. `100`

Answer

To solve the problem step by step, we need to find the first term of a geometric progression (G.P.) given that it is 2 more than the second term and the sum of the infinite series is 50.

### Step 1: Define the terms
Let:
- \( A \) = first term of the G.P.
- \( R \) = common ratio of the G.P.

The terms of the G.P. are:
- First term: \( A \)
- Second term: \( AR \)

### Step 2: Set up the equation based on the problem statement
According to the problem, the first term is 2 more than the second term:
\[ A = AR + 2 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ A - AR = 2 \]
Factoring out \( A \) from the left side:
\[ A(1 - R) = 2 \quad \text{(Equation 1)} \]

### Step 4: Use the sum of the infinite G.P.
The sum of an infinite G.P. is given by the formula:
\[ S_{\infty} = \frac{A}{1 - R} \]
According to the problem, this sum is equal to 50:
\[ \frac{A}{1 - R} = 50 \]
Rearranging gives:
\[ A = 50(1 - R) \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Substituting \( A \) from Equation 2 into Equation 1:
\[ 50(1 - R)(1 - R) = 2 \]
Expanding this gives:
\[ 50(1 - 2R + R^2) = 2 \]
Dividing both sides by 50:
\[ 1 - 2R + R^2 = \frac{2}{50} \]
Simplifying further:
\[ 1 - 2R + R^2 = \frac{1}{25} \]
Rearranging gives:
\[ R^2 - 2R + 1 - \frac{1}{25} = 0 \]
Multiplying through by 25 to eliminate the fraction:
\[ 25R^2 - 50R + 25 - 1 = 0 \]
This simplifies to:
\[ 25R^2 - 50R + 24 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 25, b = -50, c = 24 \):
\[ R = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 25 \cdot 24}}{2 \cdot 25} \]
Calculating the discriminant:
\[ R = \frac{50 \pm \sqrt{2500 - 2400}}{50} \]
\[ R = \frac{50 \pm \sqrt{100}}{50} \]
\[ R = \frac{50 \pm 10}{50} \]
This gives us two possible values for \( R \):
1. \( R = \frac{60}{50} = \frac{6}{5} \)
2. \( R = \frac{40}{50} = \frac{4}{5} \)

### Step 7: Find the first term \( A \)
Using \( R = \frac{4}{5} \) in Equation 2:
\[ A = 50(1 - \frac{4}{5}) = 50 \cdot \frac{1}{5} = 10 \]

Using \( R = \frac{6}{5} \):
\[ A = 50(1 - \frac{6}{5}) = 50 \cdot \left(-\frac{1}{5}\right) = -10 \]

### Conclusion
The first term of the G.P. can be either \( 10 \) or \( -10 \).

### Final Answer
The first term of the G.P. is \( \pm 10 \).

---