MCQ
If $x,\,2x + 2,\,3x + 3,$are in $G.P.$, then the fourth term is
- A$27$
- B$- 27$
- C$13.5$
- ✓$- 13.5$
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 ${T_4} = \frac{8}{{16}}{( - 3)^3} = - \frac{{27}}{2} = - 13.5$.
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