Let a, b and c be in GP with common ratio, r where `a !=0 and 0 lt r le (1)/(2)`. If 3a, 7b and 15 c are the first three terms of an AP, then the 4th term of this AP is

Options

A. 5a
B. `(2)/(3) a`
C. a
D. `(7)/(3)a`

Explanation

To solve the problem step by step, we will follow the logical flow of the information provided.

### Given:
- \( a, b, c \) are in GP with common ratio \( r \).
- \( a \neq 0 \) and \( 0 < r \leq \frac{1}{2} \).
- The terms \( 3a, 7b, 15c \) are the first three terms of an AP.

### Step 1: Express \( b \) and \( c \) in terms of \( a \)
Since \( a, b, c \) are in GP, we can express \( b \) and \( c \) as:
- \( b = ar \)
- \( c = ar^2 \)

### Step 2: Set up the condition for AP
For three terms \( 3a, 7b, 15c \) to be in AP, the condition is:
\[ 2(7b) = 3a + 15c \]

### Step 3: Substitute \( b \) and \( c \)
Substituting \( b \) and \( c \) into the equation:
\[ 2(7(ar)) = 3a + 15(ar^2) \]
This simplifies to:
\[ 14ar = 3a + 15ar^2 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 15ar^2 - 14ar + 3a = 0 \]
Factoring out \( a \) (since \( a \neq 0 \)):
\[ 15r^2 - 14r + 3 = 0 \]

### Step 5: Solve the quadratic equation
Now we will solve the quadratic equation \( 15r^2 - 14r + 3 = 0 \) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 15, b = -14, c = 3 \):
\[ r = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 15 \cdot 3}}{2 \cdot 15} \]
\[ r = \frac{14 \pm \sqrt{196 - 180}}{30} \]
\[ r = \frac{14 \pm \sqrt{16}}{30} \]
\[ r = \frac{14 \pm 4}{30} \]

Calculating the two possible values for \( r \):
1. \( r = \frac{18}{30} = \frac{3}{5} \) (not valid since \( r \leq \frac{1}{2} \))
2. \( r = \frac{10}{30} = \frac{1}{3} \) (valid)

### Step 6: Find the common difference \( D \)
Now we need to find the common difference \( D \) of the AP:
\[ D = 7b - 3a \]
Substituting \( b = ar \):
\[ D = 7(ar) - 3a = 7a\left(\frac{1}{3}\right) - 3a = \frac{7a}{3} - 3a \]
\[ D = \frac{7a - 9a}{3} = \frac{-2a}{3} \]

### Step 7: Find the fourth term of the AP
The fourth term of the AP can be found using:
\[ T_4 = T_1 + 3D \]
Where \( T_1 = 3a \):
\[ T_4 = 3a + 3\left(\frac{-2a}{3}\right) \]
\[ T_4 = 3a - 2a = a \]

### Final Answer:
The fourth term of the AP is \( a \).

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