A) 13.5
B) \[-13.5\]
C) 27
D) \[-27\]
Correct Answer: B
Solution :
Given,\[x,2x+2,3x+3,...\]are in GP. \[\therefore \]\[{{(2x+2)}^{2}}=x(3x+3)\] \[[\because {{b}^{2}}=ac]\] \[\Rightarrow \] \[4{{x}^{2}}+4+8x=3{{x}^{2}}+3x\] \[\Rightarrow \] \[{{x}^{2}}+5x+4=0\] \[\Rightarrow \] \[{{x}^{2}}+4x+x+4=0\] \[\Rightarrow \] \[(x+4)(x+1)=0\] \[\Rightarrow \] \[x=-4,x\ne -1\] \[\Rightarrow \] \[x=-4\] Common ratio, \[r=\frac{2x+2}{x}=\frac{-8+2}{-4}=\frac{-6}{-4}=\frac{3}{2}\] \[\therefore \]Fourth term \[=x.{{r}^{3}}=(-4){{\left( \frac{3}{2} \right)}^{3}}=-4.\frac{27}{8}\] \[=-\frac{27}{2}=-13.5\]
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