A) 27
B) \[-27\]
C) 13.5
D) \[-13.5\]
Correct Answer: D
Solution :
Given that \[x,\ 2x+2,\ 3x+3\] are in G.P. 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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