The product of `3^(rd)` and `8^(th)` terms of a G.P. is 243. If its `4^(th)` term is 3, find its `7^(th)` term.

Answer

To solve the problem step by step, we will use the properties of a Geometric Progression (G.P.).

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \).

### Step 2: Write the expressions for the terms
The \( n^{th} \) term of a G.P. is given by:
\[ T_n = a r^{n-1} \]
Thus, we can express the third and eighth terms as follows:
- \( T_3 = a r^{2} \)
- \( T_8 = a r^{7} \)

### Step 3: Set up the equation for the product of the third and eighth terms
According to the problem, the product of the third and eighth terms is 243:
\[ T_3 \times T_8 = (a r^{2}) \times (a r^{7}) = a^2 r^{9} = 243 \]
Let’s label this as Equation (1):
\[ a^2 r^{9} = 243 \]

### Step 4: Write the expression for the fourth term
The fourth term is given as:
\[ T_4 = a r^{3} = 3 \]
Let’s label this as Equation (2):
\[ a r^{3} = 3 \]

### Step 5: Solve for \( a \) in terms of \( r \)
From Equation (2), we can express \( a \) as:
\[ a = \frac{3}{r^{3}} \]

### Step 6: Substitute \( a \) into Equation (1)
Now, substitute \( a \) from Equation (2) into Equation (1):
\[ \left(\frac{3}{r^{3}}\right)^2 r^{9} = 243 \]
This simplifies to:
\[ \frac{9}{r^{6}} r^{9} = 243 \]
\[ \frac{9 r^{9}}{r^{6}} = 243 \]
\[ 9 r^{3} = 243 \]

### Step 7: Solve for \( r \)
Dividing both sides by 9:
\[ r^{3} = 27 \]
Taking the cube root:
\[ r = 3 \]

### Step 8: Substitute \( r \) back to find \( a \)
Now, substitute \( r = 3 \) back into the expression for \( a \):
\[ a = \frac{3}{3^{3}} = \frac{3}{27} = \frac{1}{9} \]

### Step 9: Find the seventh term
Now we can find the seventh term \( T_7 \):
\[ T_7 = a r^{6} = \left(\frac{1}{9}\right) \times (3^{6}) \]
Calculating \( 3^{6} \):
\[ 3^{6} = 729 \]
Thus,
\[ T_7 = \frac{729}{9} = 81 \]

### Final Answer
The seventh term \( T_7 \) is \( 81 \).

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