In a geometric progression the ratio of the sum of the first 5 terms to the sum of their reciprocals is 49 and sum of the first and the third term is 35. The fifth term of the G.P. is

Options

A. 7
B. `7//2`
C. `7//4`
D. `7//8`

Answer

To solve the problem step by step, we will follow the information provided in the question regarding the geometric progression (G.P.).

### 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 five terms of the G.P. are:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)

### Step 2: Calculate the sum of the first 5 terms
The sum of the first 5 terms \( S_1 \) can be calculated using the formula for the sum of a geometric series:
\[ S_1 = a + ar + ar^2 + ar^3 + ar^4 = a(1 + r + r^2 + r^3 + r^4) \]
Using the formula for the sum of a geometric series, we can simplify this to:
\[ S_1 = a \frac{r^5 - 1}{r - 1} \quad \text{(for } r \neq 1\text{)} \]

### Step 3: Calculate the sum of the reciprocals of the first 5 terms
The sum of the reciprocals of the first 5 terms \( S_2 \) is:
\[ S_2 = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4} \]
Factoring out \( \frac{1}{a} \):
\[ S_2 = \frac{1}{a} \left( 1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4} \right) = \frac{1}{a} \frac{1 - \frac{1}{r^5}}{1 - \frac{1}{r}} = \frac{1}{a} \cdot \frac{r^5 - 1}{r^4(r - 1)} \]

### Step 4: Set up the ratio of the sums
According to the problem, the ratio of the sum of the first 5 terms to the sum of their reciprocals is 49:
\[ \frac{S_1}{S_2} = 49 \]
Substituting the expressions for \( S_1 \) and \( S_2 \):
\[ \frac{a \frac{r^5 - 1}{r - 1}}{\frac{1}{a} \cdot \frac{r^5 - 1}{r^4(r - 1)}} = 49 \]
This simplifies to:
\[ a^2 r^4 = 49 \]

### Step 5: Use the second condition
We are also given that the sum of the first and third terms is 35:
\[ a + ar^2 = 35 \]
Factoring out \( a \):
\[ a(1 + r^2) = 35 \]

### Step 6: Solve the equations
Now we have two equations:
1. \( a^2 r^4 = 49 \)
2. \( a(1 + r^2) = 35 \)

From the second equation, we can express \( a \):
\[ a = \frac{35}{1 + r^2} \]

Substituting this into the first equation:
\[ \left(\frac{35}{1 + r^2}\right)^2 r^4 = 49 \]
This simplifies to:
\[ \frac{1225 r^4}{(1 + r^2)^2} = 49 \]
Cross-multiplying gives:
\[ 1225 r^4 = 49(1 + r^2)^2 \]
Expanding the right side:
\[ 1225 r^4 = 49(1 + 2r^2 + r^4) \]
Rearranging gives:
\[ 1225 r^4 - 49r^4 - 98r^2 - 49 = 0 \]
This simplifies to:
\[ 1176 r^4 - 98r^2 - 49 = 0 \]

### Step 7: Let \( x = r^2 \)
Let \( x = r^2 \):
\[ 1176 x^2 - 98 x - 49 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{98 \pm \sqrt{(-98)^2 - 4 \cdot 1176 \cdot (-49)}}{2 \cdot 1176} \]
Calculating the discriminant:
\[ = 9604 + 230976 = 240580 \]
Calculating the roots:
\[ x = \frac{98 \pm \sqrt{240580}}{2352} \]
Calculating \( r^2 \) gives us the value for \( r \).

### Step 8: Find the fifth term
The fifth term of the G.P. is given by:
\[ ar^4 \]
Using the values of \( a \) and \( r \) obtained from the previous steps, we can calculate the fifth term.