A) 6
B) 7
C) 8
D) 9
Correct Answer: D
Solution :
\[{{T}_{r\,}}={{\,}^{14}}{{C}_{r-1}}\,{{x}^{r-1}}\,;\,\,{{T}_{r+1}}\,=\,{{\,}^{14}}{{C}_{r}}\,{{x}^{r}}\]; \[{{T}_{r+2}}={{\,}^{14}}{{C}_{r+1}}\,{{x}^{r+1}}\] By the given condition \[2\,{{.}^{14}}{{C}_{r}}={{\,}^{14}}{{C}_{r-1}}+{{\,}^{14}}{{C}_{r+1}}\] ?..(i) Þ \[2\,.\frac{14!}{r!\,\,(14-r)!}\,=\,\frac{14!}{(r-1)!\,\,(15-r)!}+\frac{14!}{(r+1)!\,(13-r)!}\] Þ \[\frac{2}{r.(r-1)!.(14-r).(13-r)!}\] =\[\frac{1}{(r-1)!.(15-r).(14-r).(13-r)!}\]+\[\frac{1}{(r+1)\,r(r-1)\,!\,(13-r)\,!}\] Þ \[\frac{2}{r(14-r)}=\frac{1}{(15-r)(14-r)}+\frac{1}{(r+1)r}\] Þ \[\frac{1}{r(14-r)}-\frac{1}{(15-r)(14-r)}=\frac{1}{(r+1)r}-\frac{1}{r(14-r)}\] Þ \[\frac{(15-r)-r}{r(15-r)(14-r)}=\frac{(14-r)-(r+1)}{(r+1)r(14-r)}\] Þ \[\frac{15-2r}{15-r}=\frac{13-2r}{r+1}\] Þ \[15r+15-2{{r}^{2}}-2r=195-30r-13r+2{{r}^{2}}\] Þ \[4{{r}^{2}}-56r+180=0\,\,\Rightarrow \,{{r}^{2}}-14r+45=0\] Þ \[(r-5)(r-9)=0\Rightarrow r=5,\,9\] But 5 is not given. Hence r = 9. Trick: Put the value of r from options in equation (i), only satisfy it.
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