A) 6
B) 4
C) 8
D) 10
Correct Answer: C
Solution :
Let the three consecutive coefficients be \[^{n}{{C}_{r-1}}=28,{{\,}^{n}}{{C}_{r}}=56\]and \[^{n}{{C}_{r+1}}=70,\]so that \[\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=\frac{n-r+1}{r}=\frac{56}{28}=2\] and \[\frac{^{n}{{C}_{r+1}}}{^{n}{{C}_{r}}}=\frac{n-r}{r+1}=\frac{70}{56}=\frac{5}{4}\] This gives \[n+1=3r\]and \[4n-5=9r\] \ \[\frac{4n-5}{n+1}=3\Rightarrow n=8\]
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