A) 4
B) 5
C) 6
D) 7
E) 8
Correct Answer: E
Solution :
Given, \[{{T}_{n+1}}-{{T}_{n}}=28\] \[\Rightarrow \] \[^{n+1}{{C}_{3}}{{-}^{n}}{{C}_{3}}=28\] \[\Rightarrow \] \[\frac{(n+1)!}{3!(n-2)!}-\frac{n!}{(n-3)!3!}=28\] \[\Rightarrow \]\[\frac{1}{6}[(n+1)(n)(n-1)-n(n-1)(n-2)]=28\] \[\Rightarrow \] \[n[{{n}^{2}}-1-({{n}^{2}}-3n+2)]=168\] \[\Rightarrow \] \[{{n}^{2}}-n-56=0\] \[\Rightarrow \] \[n=8\] \[(n\ne -7)\]
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