In an infinite G.P. the sum of first three terms is 70. If the extreme terms are multiplied by 4 and the middle term is multiplied by 5, the resulting terms form an A.P. then the sum to infinite terms of G.P. is :

Options

A. 120
B. 40
C. 160
D. 80

Answer

To solve the problem step-by-step, we will follow the logic presented in the video transcript and derive the solution systematically.

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( A \) and the common ratio be \( R \). The first three terms of the G.P. are:
- First term: \( A \)
- Second term: \( AR \)
- Third term: \( AR^2 \)

### Step 2: Set up the equation for the sum of the first three terms
According to the problem, the sum of the first three terms is given as:
\[ A + AR + AR^2 = 70 \]
Factoring out \( A \) gives:
\[ A(1 + R + R^2) = 70 \quad \text{(1)} \]

### Step 3: Modify the terms as per the problem statement
The problem states that if the extreme terms (first and third terms) are multiplied by 4, and the middle term is multiplied by 5, the resulting terms form an A.P. Therefore, we have:
- First term: \( 4A \)
- Second term: \( 5AR \)
- Third term: \( 4AR^2 \)

### Step 4: Set up the condition for A.P.
For the three terms \( 4A, 5AR, 4AR^2 \) to be in A.P., the following condition must hold:
\[ 2 \times (5AR) = (4A + 4AR^2) \]
Simplifying this gives:
\[ 10AR = 4A + 4AR^2 \]
Rearranging leads to:
\[ 10AR - 4AR^2 - 4A = 0 \]
Factoring out \( 2A \) (assuming \( A \neq 0 \)):
\[ 2A(5R - 2R^2 - 2) = 0 \]
Thus, we have:
\[ 5R - 2R^2 - 2 = 0 \quad \text{(2)} \]

### Step 5: Solve the quadratic equation
Rearranging equation (2):
\[ 2R^2 - 5R + 2 = 0 \]
Using the quadratic formula \( R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -5, c = 2 \):
\[ R = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
Calculating the roots:
1. \( R = \frac{8}{4} = 2 \)
2. \( R = \frac{2}{4} = \frac{1}{2} \)

### Step 6: Determine the valid value of \( R \)
Since we are dealing with an infinite G.P., the common ratio \( R \) must be less than 1. Therefore, we take:
\[ R = \frac{1}{2} \]

### Step 7: Substitute \( R \) back to find \( A \)
Substituting \( R = \frac{1}{2} \) back into equation (1):
\[ A(1 + \frac{1}{2} + \left(\frac{1}{2}\right)^2) = 70 \]
Calculating \( 1 + \frac{1}{2} + \frac{1}{4} = \frac{4}{4} + \frac{2}{4} + \frac{1}{4} = \frac{7}{4} \):
\[ A \cdot \frac{7}{4} = 70 \]
Thus,
\[ A = 70 \cdot \frac{4}{7} = 40 \]

### Step 8: Calculate the sum of the infinite G.P.
The sum \( S \) of an infinite G.P. is given by:
\[ S = \frac{A}{1 - R} \]
Substituting the values of \( A \) and \( R \):
\[ S = \frac{40}{1 - \frac{1}{2}} = \frac{40}{\frac{1}{2}} = 40 \cdot 2 = 80 \]

### Final Answer
The sum to infinite terms of the G.P. is:
\[ \boxed{80} \]