In an infinite G.P. the sum of first three terms is 70. If the externme terms are multipled 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, let's denote the first term of the infinite geometric progression (G.P.) as \( a \) and the common ratio as \( r \).

### Step 1: Set up the equation for the sum of the first three terms
The sum of the first three terms of a G.P. is given by:
\[ S_3 = a + ar + ar^2 = a(1 + r + r^2) \]
According to the problem, this sum equals 70:
\[ a(1 + r + r^2) = 70 \quad \text{(Equation 1)} \]

### Step 2: Formulate the new terms after multiplication
The extreme terms are multiplied by 4 and the middle term is multiplied by 5, resulting in the new terms:
- First term: \( 4a \)
- Middle term: \( 5ar \)
- Last term: \( 4ar^2 \)

These terms must form an arithmetic progression (A.P.). For three terms \( x, y, z \) to be in A.P., the condition is:
\[ 2y = x + z \]
Substituting our terms:
\[ 2(5ar) = 4a + 4ar^2 \]
This simplifies to:
\[ 10ar = 4a + 4ar^2 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ 10ar - 4ar^2 - 4a = 0 \]
Factoring out \( 2a \):
\[ 2a(5r - 2r^2 - 2) = 0 \]
Since \( a \neq 0 \), we can divide by \( 2a \):
\[ 5r - 2r^2 - 2 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the quadratic equation
Rearranging Equation 2 gives us:
\[ 2r^2 - 5r + 2 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2, b = -5, c = 2 \):
\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
This gives us two possible values for \( r \):
\[ r_1 = \frac{8}{4} = 2 \quad \text{and} \quad r_2 = \frac{2}{4} = \frac{1}{2} \]

### Step 5: Find the corresponding values of \( a \)
Now, we substitute these values of \( r \) back into Equation 1 to find \( a \).

1. For \( r = 2 \):
\[ a(1 + 2 + 4) = 70 \implies a \cdot 7 = 70 \implies a = 10 \]

2. For \( r = \frac{1}{2} \):
\[ a(1 + \frac{1}{2} + \frac{1}{4}) = 70 \implies a \cdot \frac{7}{4} = 70 \implies a = 40 \]

### Step 6: Calculate the sum of infinite terms of the G.P.
The sum of an infinite G.P. is given by:
\[ S_{\infty} = \frac{a}{1 - r} \]
We need to consider only the value of \( r \) that is between 0 and 1, which is \( r = \frac{1}{2} \) with \( a = 40 \):
\[ S_{\infty} = \frac{40}{1 - \frac{1}{2}} = \frac{40}{\frac{1}{2}} = 80 \]

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