The first term of a G.P. is -3. If the 4th term of this G.P. is the square of the 2nd term, then find its 7th term.

Explanation

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

### Step 1: Identify the first term and the formula for the nth term of a G.P.
The first term of the G.P. is given as:
\[ a = -3 \]

The formula for the nth term of a G.P. is:
\[ a_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

### Step 2: Write the expressions for the 2nd and 4th terms.
The 2nd term \( a_2 \) is:
\[ a_2 = a \cdot r^{2-1} = a \cdot r = -3r \]

The 4th term \( a_4 \) is:
\[ a_4 = a \cdot r^{4-1} = a \cdot r^3 = -3r^3 \]

### Step 3: Set up the equation based on the given condition.
According to the problem, the 4th term is equal to the square of the 2nd term:
\[ a_4 = (a_2)^2 \]
Substituting the expressions we found:
\[ -3r^3 = (-3r)^2 \]

### Step 4: Simplify the equation.
The right side simplifies to:
\[ (-3r)^2 = 9r^2 \]
So, we have:
\[ -3r^3 = 9r^2 \]

### Step 5: Rearrange the equation.
To solve for \( r \), we can rearrange the equation:
\[ -3r^3 - 9r^2 = 0 \]
Factoring out \( -3r^2 \):
\[ -3r^2(r + 3) = 0 \]

### Step 6: Solve for \( r \).
Setting each factor to zero gives us:
1. \( -3r^2 = 0 \) → \( r = 0 \) (not valid for a G.P.)
2. \( r + 3 = 0 \) → \( r = -3 \)

### Step 7: Find the 7th term of the G.P.
Now that we have \( r = -3 \), we can find the 7th term \( a_7 \):
\[ a_7 = a \cdot r^{7-1} = a \cdot r^6 \]
Substituting the values:
\[ a_7 = -3 \cdot (-3)^6 \]

### Step 8: Calculate \( (-3)^6 \).
Calculating \( (-3)^6 \):
\[ (-3)^6 = 729 \]
Thus,
\[ a_7 = -3 \cdot 729 = -2187 \]

### Final Answer:
The 7th term of the G.P. is:
\[ a_7 = -2187 \]

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