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

Options

A. 6
B. 14
C. `27//14`
D. `57//15`

Answer

To find the 7th term of the geometric progression (GP), we can follow these steps:

### Step 1: Understand the terms of the GP
In a GP, the nth term can be expressed as:
\[ T_n = A \cdot R^{(n-1)} \]
where \( A \) is the first term, \( R \) is the common ratio, and \( n \) is the term number.

### Step 2: Write down the equations for the given terms
From the problem, we know:
- The 10th term \( T_{10} = 9 \) can be expressed as:
\[ T_{10} = A \cdot R^{9} = 9 \]
(Equation 1)

- The 4th term \( T_{4} = 4 \) can be expressed as:
\[ T_{4} = A \cdot R^{3} = 4 \]
(Equation 2)

### Step 3: Divide Equation 1 by Equation 2
To eliminate \( A \) and find \( R \), we can divide Equation 1 by Equation 2:
\[ \frac{A \cdot R^{9}}{A \cdot R^{3}} = \frac{9}{4} \]
This simplifies to:
\[ R^{6} = \frac{9}{4} \]

### Step 4: Solve for \( R \)
Taking the sixth root of both sides:
\[ R = \left(\frac{9}{4}\right)^{\frac{1}{6}} = \frac{3^{1/3}}{2^{1/3}} = \frac{3^{1/3}}{2^{1/3}} = \frac{3^{1/3}}{2^{1/3}} = \frac{3^{1/3}}{2^{1/3}} \]

### Step 5: Substitute \( R \) back into Equation 2 to find \( A \)
Now, substitute \( R \) back into Equation 2 to find \( A \):
\[ A \cdot R^{3} = 4 \]
Substituting \( R^{3} \):
\[ A \cdot \left(\frac{9}{4}\right)^{\frac{1}{2}} = 4 \]
This simplifies to:
\[ A \cdot \frac{3}{2} = 4 \]
Thus:
\[ A = \frac{4 \cdot 2}{3} = \frac{8}{3} \]

### Step 6: Find the 7th term
Now we can find the 7th term \( T_{7} \):
\[ T_{7} = A \cdot R^{6} \]
Substituting the values of \( A \) and \( R^{6} \):
\[ T_{7} = \frac{8}{3} \cdot \frac{9}{4} = \frac{8 \cdot 9}{3 \cdot 4} = \frac{72}{12} = 6 \]

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

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