If for a G.P., `S_(2)` =8 and `S_(4)`=80 , then : (a,r) `-=`

Options

A. (2,4)
B. (2,3)
C. (3,4)
D. none of these

Answer

To solve the problem, we need to find the values of \( A \) (the first term) and \( r \) (the common ratio) of a geometric progression (G.P.) given that \( S_2 = 8 \) and \( S_4 = 80 \).

### Step 1: Write the formulas for \( S_2 \) and \( S_4 \)

The sum of the first \( n \) terms of a geometric progression is given by the formula:

\[ S_n = \frac{A(r^n - 1)}{r - 1} \]

For \( S_2 \):

\[ S_2 = \frac{A(r^2 - 1)}{r - 1} = 8 \]

For \( S_4 \):

\[ S_4 = \frac{A(r^4 - 1)}{r - 1} = 80 \]

### Step 2: Set up the equations

From the above, we have two equations:

1. \(\frac{A(r^2 - 1)}{r - 1} = 8\) (Equation 1)
2. \(\frac{A(r^4 - 1)}{r - 1} = 80\) (Equation 2)

### Step 3: Simplify Equation 2

We can express \( r^4 - 1 \) in terms of \( r^2 \):

\[ r^4 - 1 = (r^2 - 1)(r^2 + 1) \]

Substituting this into Equation 2 gives:

\[ \frac{A((r^2 - 1)(r^2 + 1))}{r - 1} = 80 \]

### Step 4: Divide Equation 2 by Equation 1

Now, divide Equation 2 by Equation 1:

\[ \frac{\frac{A((r^2 - 1)(r^2 + 1))}{r - 1}}{\frac{A(r^2 - 1)}{r - 1}} = \frac{80}{8} \]

This simplifies to:

\[ r^2 + 1 = 10 \]

### Step 5: Solve for \( r^2 \)

From the equation \( r^2 + 1 = 10 \):

\[ r^2 = 10 - 1 = 9 \]

Taking the square root gives:

\[ r = 3 \quad (\text{since } r \text{ must be positive}) \]

### Step 6: Substitute \( r \) back to find \( A \)

Now substitute \( r = 3 \) back into Equation 1 to find \( A \):

\[ \frac{A(3^2 - 1)}{3 - 1} = 8 \]

This simplifies to:

\[ \frac{A(9 - 1)}{2} = 8 \]

\[ \frac{8A}{2} = 8 \]

\[ 4A = 8 \implies A = 2 \]

### Final Answer

Thus, the values are:

\[ (A, r) = (2, 3) \]

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