The sum of the first and the third term of G.P. is 15 and that of the 5th and the 7th terms is 240. Find the 9th term :

Options

A. 678
B. 786
C. 867
D. 768

Answer

To solve the problem step by step, let's denote the first term of the geometric progression (G.P.) as \( a \) and the common ratio as \( r \).

### Step 1: Set up the equations based on the given information.
1. The sum of the first and third terms of the G.P. is given as:
\[ T_1 + T_3 = a + ar^2 = 15 \]
This can be rearranged to:
\[ a(1 + r^2) = 15 \quad \text{(Equation 1)} \]

2. The sum of the fifth and seventh terms of the G.P. is given as:
\[ T_5 + T_7 = ar^4 + ar^6 = 240 \]
This can be rearranged to:
\[ ar^4(1 + r^2) = 240 \quad \text{(Equation 2)} \]

### Step 2: Divide Equation 2 by Equation 1.
We can eliminate \( a(1 + r^2) \) by dividing Equation 2 by Equation 1:
\[ \frac{ar^4(1 + r^2)}{a(1 + r^2)} = \frac{240}{15} \]
This simplifies to:
\[ r^4 = 16 \]

### Step 3: Solve for \( r \).
Taking the fourth root of both sides gives:
\[ r = 2 \quad \text{(since \( r \) must be positive in this context)} \]

### Step 4: Substitute \( r \) back into Equation 1 to find \( a \).
Now substitute \( r = 2 \) back into Equation 1:
\[ a(1 + 2^2) = 15 \]
This simplifies to:
\[ a(1 + 4) = 15 \implies 5a = 15 \implies a = 3 \]

### Step 5: Find the 9th term of the G.P.
The 9th term \( T_9 \) is given by:
\[ T_9 = a \cdot r^{9-1} = a \cdot r^8 \]
Substituting the values of \( a \) and \( r \):
\[ T_9 = 3 \cdot 2^8 \]
Calculating \( 2^8 \):
\[ 2^8 = 256 \]
Thus,
\[ T_9 = 3 \cdot 256 = 768 \]

### Final Answer
The 9th term of the G.P. is \( \boxed{768} \).