The fourth term of a G.P. is square of its 2nd term and the first term is -3. Determine the
(i) 7th term
(ii) 6th term

Answer

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

### Given:
1. The first term \( a_1 = -3 \)
2. The fourth term \( a_4 \) is the square of the second term \( a_2 \).

### Step 1: Write the formulas for the terms of the G.P.
The \( n \)-th term of a G.P. can be expressed as:
\[ a_n = a_1 \cdot r^{n-1} \]
Where \( a_1 \) is the first term and \( r \) is the common ratio.

### Step 2: Express the terms in terms of \( a_1 \) and \( r \)
- The second term \( a_2 \) is:
\[ a_2 = a_1 \cdot r = -3 \cdot r \]
- The fourth term \( a_4 \) is:
\[ a_4 = a_1 \cdot r^3 = -3 \cdot r^3 \]

### Step 3: Set up the equation based on the given condition
According to the problem, the fourth term is the square of the second term:
\[ a_4 = (a_2)^2 \]
Substituting the expressions we found:
\[ -3 \cdot r^3 = (-3 \cdot r)^2 \]
This simplifies to:
\[ -3 \cdot r^3 = 9 \cdot r^2 \]

### Step 4: Rearranging the equation
We can rearrange the equation:
\[ -3r^3 - 9r^2 = 0 \]
Factoring out \(-3r^2\):
\[ -3r^2(r + 3) = 0 \]

### Step 5: Solve for \( r \)
Setting each factor to zero gives us:
1. \(-3r^2 = 0\) → \( r = 0 \) (not valid for G.P.)
2. \(r + 3 = 0\) → \( r = -3 \)

### Step 6: Find the 7th term \( a_7 \)
Using the formula for the \( n \)-th term:
\[ a_7 = a_1 \cdot r^{6} = -3 \cdot (-3)^{6} \]
Calculating \( (-3)^{6} = 729 \):
\[ a_7 = -3 \cdot 729 = -2187 \]

### Step 7: Find the 6th term \( a_6 \)
Using the formula for the 6th term:
\[ a_6 = a_1 \cdot r^{5} = -3 \cdot (-3)^{5} \]
Calculating \( (-3)^{5} = -243 \):
\[ a_6 = -3 \cdot (-243) = 729 \]

### Final Answers:
(i) The 7th term \( a_7 = -2187 \)

(ii) The 6th term \( a_6 = 729 \)

---