A) \[27.5\]
B) \[4x+5\]
C) \[-13.5\]
D) \[4x+4\]
Correct Answer: C
Solution :
Since, given series is in GP. \[\therefore \] \[{{(2x+2)}^{2}}=x\times (3x+3)\] \[\Rightarrow \] \[4{{x}^{2}}+4+8x=3{{x}^{2}}+3x\] \[\Rightarrow \] \[{{x}^{2}}+5x+4=0\] \[\Rightarrow \] \[x=\frac{-5\pm \sqrt{25-16}}{2}\] \[\Rightarrow \] \[x=\frac{-5\pm 3}{2}\] \[\Rightarrow \] \[x=-\frac{8}{2}=-4\] and \[x=-\frac{2}{2}=-1\] At \[x=-1,\] second terms become zero so we neglect that. At \[x=-4,\,\,a=-4,\,\,\,r=3/2\] \[\therefore \]\[{{T}_{4}}=-4\times {{(3/2)}^{3}}=-\frac{27}{2}=-13.5\]
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