If the second term of a G.P. is 2 and the sum of its infinite terms is 8, then G.P. is
(i) `8,2,1/2,1/8,`.
(ii) `10,2,2/5,2/25`,.
(iii) `4,2,1,1/2,1/4`.
(iv) `6,3,3/2,3/4`,…..

Options

A. `8,2,1/2,1/8,`.
B. `10,2,2/5,2/25`,.
C. `4,2,1,1/2,1/4`.
D. `6,3,3/2,3/4`,…..

Answer

To solve the problem, we need to find the geometric progression (G.P.) given that the second term is 2 and the sum of its infinite terms is 8. Let's break this down step by step.

### 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 terms of the G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- And so on...

### Step 2: Use the information about the second term
According to the problem, the second term is given as 2:
\[ ar = 2 \quad \text{(Equation 1)} \]

### Step 3: Use the information about the sum of infinite terms
The sum \( S \) of an infinite G.P. is given by the formula:
\[ S = \frac{a}{1 - r} \]
We know the sum is 8:
\[ \frac{a}{1 - r} = 8 \quad \text{(Equation 2)} \]

### Step 4: Solve for \( a \) in terms of \( r \)
From Equation 2, we can express \( a \) in terms of \( r \):
\[ a = 8(1 - r) \]

### Step 5: Substitute \( a \) in Equation 1
Now, substitute the expression for \( a \) from Equation 2 into Equation 1:
\[ (8(1 - r))r = 2 \]
Expanding this gives:
\[ 8r - 8r^2 = 2 \]
Rearranging the equation:
\[ 8r^2 - 8r + 2 = 0 \]

### Step 6: Simplify the quadratic equation
Dividing the entire equation by 2:
\[ 4r^2 - 4r + 1 = 0 \]

### Step 7: Factor the quadratic equation
This can be factored as:
\[ (2r - 1)^2 = 0 \]
Thus, we find:
\[ 2r - 1 = 0 \implies r = \frac{1}{2} \]

### Step 8: Find \( a \) using the value of \( r \)
Substituting \( r \) back into Equation 1 to find \( a \):
\[ a \cdot \frac{1}{2} = 2 \implies a = 2 \cdot 2 = 4 \]

### Step 9: Write the G.P.
Now we can write the G.P. using the values of \( a \) and \( r \):
- First term: \( a = 4 \)
- Second term: \( ar = 4 \cdot \frac{1}{2} = 2 \)
- Third term: \( ar^2 = 4 \cdot \left(\frac{1}{2}\right)^2 = 1 \)
- Fourth term: \( ar^3 = 4 \cdot \left(\frac{1}{2}\right)^3 = \frac{1}{2} \)
- Fifth term: \( ar^4 = 4 \cdot \left(\frac{1}{2}\right)^4 = \frac{1}{4} \)

Thus, the G.P. is:
\[ 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots \]

### Conclusion
The correct answer is option (iii) \( 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots \)

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