Answer :
We are given the first three terms of a geometric sequence:
[tex]$$
a_1 = x+3, \quad a_2=-2x^2-6x, \quad a_3=4x^3+12x^2.
$$[/tex]
Step 1. Find the common ratio.
For a geometric sequence, the common ratio [tex]$r$[/tex] is found by dividing any term by its preceding term. Using the first two terms, we have
[tex]$$
r = \frac{a_2}{a_1} = \frac{-2x^2-6x}{x+3}.
$$[/tex]
Notice that the numerator can be factored:
[tex]$$
-2x^2-6x = -2x(x+3).
$$[/tex]
Thus,
[tex]$$
r = \frac{-2x(x+3)}{x+3}.
$$[/tex]
Provided [tex]$x+3\neq 0$[/tex], we cancel the common factor:
[tex]$$
r = -2x.
$$[/tex]
For confirmation, you can also check with the third term:
[tex]$$
r = \frac{a_3}{a_2} = \frac{4x^3+12x^2}{-2x^2-6x}.
$$[/tex]
Factor both numerator and denominator:
- Numerator: [tex]$4x^3+12x^2 = 4x^2(x+3)$[/tex].
- Denominator: [tex]$-2x^2-6x = -2x(x+3)$[/tex].
Then,
[tex]$$
r = \frac{4x^2(x+3)}{-2x(x+3)} = -2x.
$$[/tex]
Both methods confirm that
[tex]$$
r = -2x.
$$[/tex]
Step 2. Use the formula for the [tex]$n$[/tex]th term.
The [tex]$n$[/tex]th term of a geometric sequence is given by
[tex]$$
a_n = a_1 \cdot r^{n-1}.
$$[/tex]
For the eighth term ([tex]$n=8$[/tex]):
[tex]$$
a_8 = a_1 \cdot r^{7} = (x+3) \cdot (-2x)^7.
$$[/tex]
Step 3. Compute [tex]$(-2x)^7$[/tex].
We have
[tex]$$
(-2x)^7 = (-2)^7 \cdot x^7 = -128 \cdot x^7.
$$[/tex]
Step 4. Multiply by the first term.
Substitute back into the expression for [tex]$a_8$[/tex]:
[tex]$$
a_8 = (x+3) \cdot (-128x^7) = -128x^7(x+3).
$$[/tex]
Distribute [tex]$-128x^7$[/tex] over [tex]$(x+3)$[/tex]:
[tex]$$
a_8 = -128x^8 - 384x^7.
$$[/tex]
Thus, the eighth term of the sequence is
[tex]$$
\boxed{-128x^8-384x^7}.
$$[/tex]
[tex]$$
a_1 = x+3, \quad a_2=-2x^2-6x, \quad a_3=4x^3+12x^2.
$$[/tex]
Step 1. Find the common ratio.
For a geometric sequence, the common ratio [tex]$r$[/tex] is found by dividing any term by its preceding term. Using the first two terms, we have
[tex]$$
r = \frac{a_2}{a_1} = \frac{-2x^2-6x}{x+3}.
$$[/tex]
Notice that the numerator can be factored:
[tex]$$
-2x^2-6x = -2x(x+3).
$$[/tex]
Thus,
[tex]$$
r = \frac{-2x(x+3)}{x+3}.
$$[/tex]
Provided [tex]$x+3\neq 0$[/tex], we cancel the common factor:
[tex]$$
r = -2x.
$$[/tex]
For confirmation, you can also check with the third term:
[tex]$$
r = \frac{a_3}{a_2} = \frac{4x^3+12x^2}{-2x^2-6x}.
$$[/tex]
Factor both numerator and denominator:
- Numerator: [tex]$4x^3+12x^2 = 4x^2(x+3)$[/tex].
- Denominator: [tex]$-2x^2-6x = -2x(x+3)$[/tex].
Then,
[tex]$$
r = \frac{4x^2(x+3)}{-2x(x+3)} = -2x.
$$[/tex]
Both methods confirm that
[tex]$$
r = -2x.
$$[/tex]
Step 2. Use the formula for the [tex]$n$[/tex]th term.
The [tex]$n$[/tex]th term of a geometric sequence is given by
[tex]$$
a_n = a_1 \cdot r^{n-1}.
$$[/tex]
For the eighth term ([tex]$n=8$[/tex]):
[tex]$$
a_8 = a_1 \cdot r^{7} = (x+3) \cdot (-2x)^7.
$$[/tex]
Step 3. Compute [tex]$(-2x)^7$[/tex].
We have
[tex]$$
(-2x)^7 = (-2)^7 \cdot x^7 = -128 \cdot x^7.
$$[/tex]
Step 4. Multiply by the first term.
Substitute back into the expression for [tex]$a_8$[/tex]:
[tex]$$
a_8 = (x+3) \cdot (-128x^7) = -128x^7(x+3).
$$[/tex]
Distribute [tex]$-128x^7$[/tex] over [tex]$(x+3)$[/tex]:
[tex]$$
a_8 = -128x^8 - 384x^7.
$$[/tex]
Thus, the eighth term of the sequence is
[tex]$$
\boxed{-128x^8-384x^7}.
$$[/tex]
