A) \[{{m}^{2}}-m\left( 4r-1 \right)+4{{r}^{2}}-2=0\]
B) \[{{m}^{2}}-m\left( 4r+1 \right)+4{{r}^{2}}+2=0\]
C) \[{{m}^{2}}-m(4r+1)+4{{r}^{2}}-2=0\]
D) \[{{m}^{2}}-m\left( 4r-1 \right)+4{{r}^{2}}+2=0\]
Correct Answer: C
Solution :
[c] Given \[^{m}{{C}_{r-1}},\,\,{{\,}^{m}}{{C}_{r}},\,\,{{\,}^{m}}{{C}_{r+1}}\] are in A.P. \[{{2}^{m}}{{C}_{r}}={{\,}^{m}}{{C}_{r-1}}+{{\,}^{m}}{{C}_{r+1}}\] \[\Rightarrow 2=\frac{^{m}{{C}_{r-1}}}{^{m}{{C}_{r}}}+\frac{^{m}{{C}_{r+1}}}{^{m}{{C}_{r}}}=\frac{r}{m-r+1}+\frac{m-r}{r+1}\] \[\Rightarrow {{m}^{2}}-m(4r+1)+4{{r}^{2}}-2=0\].
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