A) \[\frac{7}{13}\]
B) 2
C) 4
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[{{T}_{7}}=A+6d=a;{{T}_{11}}=A+10d=b;\] \[{{T}_{13}}=A+12d=c\] Now a, b, c are in G.P. \[\therefore \,\,\,\,\text{ }{{b}^{2}}=ac\] \[\Rightarrow \,\,\,\,{{\left( A+10d \right)}^{2}}=\left( A+6d \right)\left( A+12d \right)\] \[\Rightarrow \,\,\,\,{{A}^{2}}+100{{d}^{2}}+20Ad={{A}^{2}}+18Ad+72{{d}^{2}}\] \[\Rightarrow \,\,\,\,\,A+14d=0,\text{ }A=-14d\] \[\frac{a}{c}\,=\,\,\frac{A+6d}{A+12d}\,=\,\frac{-8d}{-2d}\,\,=\,\,4\]You need to login to perform this action.
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