Explanation

(c)
Since, $(2x + 2)^{2} = x \times (3x + 3)$ $\lbrack\because b^{2} = ac\rbrack$
$\Rightarrow x^{2} + 5x + 4 = 0\ \Rightarrow x = \frac{- 5 \pm \sqrt{25 - 16}}{2}$
$\Rightarrow x = \frac{- 5 \pm 3}{2}$
$\Rightarrow x = - \frac{8}{2} = - 4$ and $x = - \frac{2}{2} = - 1$
At $x = - 1$, second terms become zero, so we neglect that.
At $x = - 4,\ a = - 4,\ r = \frac{3}{2}$
$\therefore\ T_{4} = - 4 \times \left( \frac{3}{2} \right)^{3} = - \frac{27}{2} = - 13.5$