8. sequence and series — MATHS STD 11 Science — Question

Therefore, ${(2x + 2)^2} = x(3x + 3) $

$\Rightarrow {x^2} + 5x + 4 = 0$

$ \Rightarrow (x + 4)(x + 1) = 0$

$\Rightarrow x = - 1,\; - 4$

Now first term $a = x$

Second term $ar = 2(x + 1)$

$ \Rightarrow r = \frac{{2(x + 1)}}{x}$

then ${4^{th}}$ term $ = a{r^3}$$ = x{\left[ {\frac{{2(x + 1)}}{x}} \right]^3} = \frac{8}{{{x^2}}}{(x + 1)^3}$

Putting $x = - 4$

We get ${T_4} = \frac{8}{{16}}{( - 3)^3} = - \frac{{27}}{2} = - 13.5$.