If the 10th term of a GP is 9 and 4th term is 4, then what is its 7th term?

Options

A. 6
B. 14
C. `27//14`
D. `56//15`

Explanation

To find the 7th term of a geometric progression (GP) given that the 10th term is 9 and the 4th term is 4, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the terms of the GP:**
- The nth term of a GP can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
- Where \( a \) is the first term and \( r \) is the common ratio.

2. **Write the equations for the given terms:**
- For the 10th term:
\[ T_{10} = a \cdot r^{9} = 9 \quad \text{(Equation 1)} \]
- For the 4th term:
\[ T_{4} = a \cdot r^{3} = 4 \quad \text{(Equation 2)} \]

3. **Divide Equation 1 by Equation 2:**
- This will eliminate \( a \):
\[ \frac{T_{10}}{T_{4}} = \frac{a \cdot r^{9}}{a \cdot r^{3}} = \frac{9}{4} \]
- Simplifying gives:
\[ r^{6} = \frac{9}{4} \]

4. **Solve for \( r \):**
- Taking the square root of both sides:
\[ r^{3} = \frac{3}{2} \quad \text{(since } r^{6} = \left(\frac{3}{2}\right)^{2}\text{)} \]

5. **Substitute \( r^{3} \) back into Equation 2 to find \( a \):**
- From Equation 2:
\[ a \cdot r^{3} = 4 \]
- Substitute \( r^{3} = \frac{3}{2} \):
\[ a \cdot \frac{3}{2} = 4 \]
- Solving for \( a \):
\[ a = 4 \cdot \frac{2}{3} = \frac{8}{3} \]

6. **Find the 7th term \( T_{7} \):**
- Using the formula for the nth term:
\[ T_{7} = a \cdot r^{6} \]
- We already know \( a = \frac{8}{3} \) and \( r^{6} = \frac{9}{4} \):
\[ T_{7} = \frac{8}{3} \cdot \frac{9}{4} \]
- Simplifying:
\[ T_{7} = \frac{8 \cdot 9}{3 \cdot 4} = \frac{72}{12} = 6 \]

### Final Answer:
The 7th term of the GP is \( 6 \).