In a G.P. the first, third, and fifth terms may be considered as the first, fourth, and sixteenth terms of an A.P. Then the fourth term of the A.P., knowing that its first term is 5, is `10` b. `12` c. `16` d. 20

Options

A. 10
B. 12
C. 16
D. 20

Explanation

To solve the problem step by step, we need to analyze the relationship between the terms of the G.P. (Geometric Progression) and the A.P. (Arithmetic Progression).

### 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: \( a \)
- Third term: \( ar^2 \)
- Fifth term: \( ar^4 \)

### Step 2: Define the terms of the A.P.
According to the problem, the first, third, and fifth terms of the G.P. correspond to the first, fourth, and sixteenth terms of the A.P. Let the first term of the A.P. be \( A \) and the common difference be \( D \). The terms of the A.P. can be expressed as:
- First term: \( A \)
- Fourth term: \( A + 3D \)
- Sixteenth term: \( A + 15D \)

### Step 3: Set up the equations
From the problem statement, we have:
1. \( a = A \) (first term of G.P. is the first term of A.P.)
2. \( ar^2 = A + 3D \) (third term of G.P. is the fourth term of A.P.)
3. \( ar^4 = A + 15D \) (fifth term of G.P. is the sixteenth term of A.P.)

### Step 4: Substitute \( A \) with 5
We know from the problem that the first term of the A.P. is \( A = 5 \). Therefore, we can substitute \( A \) in the equations:
1. \( a = 5 \)
2. \( ar^2 = 5 + 3D \)
3. \( ar^4 = 5 + 15D \)

### Step 5: Substitute \( a \) in the equations
Substituting \( a = 5 \) into the second and third equations:
1. \( 5r^2 = 5 + 3D \)
2. \( 5r^4 = 5 + 15D \)

### Step 6: Rearranging the equations
From the second equation:
\[ 5r^2 - 5 = 3D \]
\[ D = \frac{5(r^2 - 1)}{3} \] (Equation 1)

From the third equation:
\[ 5r^4 - 5 = 15D \]
\[ D = \frac{5(r^4 - 1)}{15} = \frac{r^4 - 1}{3} \] (Equation 2)

### Step 7: Equate the two expressions for \( D \)
Setting Equation 1 equal to Equation 2:
\[ \frac{5(r^2 - 1)}{3} = \frac{r^4 - 1}{3} \]

### Step 8: Clear the denominator
Multiplying through by 3 gives:
\[ 5(r^2 - 1) = r^4 - 1 \]
\[ 5r^2 - 5 = r^4 - 1 \]
\[ r^4 - 5r^2 + 4 = 0 \]

### Step 9: Let \( x = r^2 \)
This becomes a quadratic equation:
\[ x^2 - 5x + 4 = 0 \]

### Step 10: Solve the quadratic equation
Using the quadratic formula:
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
\[ x = \frac{5 \pm \sqrt{25 - 16}}{2} \]
\[ x = \frac{5 \pm 3}{2} \]

Calculating the roots:
1. \( x = \frac{8}{2} = 4 \)
2. \( x = \frac{2}{2} = 1 \)

### Step 11: Find \( r \)
Since \( x = r^2 \):
1. \( r^2 = 4 \) implies \( r = 2 \)
2. \( r^2 = 1 \) implies \( r = 1 \)

### Step 12: Calculate \( D \)
Using \( r = 2 \):
\[ D = \frac{5(4 - 1)}{3} = \frac{15}{3} = 5 \]

### Step 13: Find the fourth term of the A.P.
The fourth term of the A.P. is:
\[ A + 3D = 5 + 3 \cdot 5 = 5 + 15 = 20 \]

### Final Answer:
The fourth term of the A.P. is \( 20 \).