A) 5
B) 7
C) 6
D) 4
Correct Answer: B
Solution :
\[{{T}_{n}}{{=}^{n}}{{C}_{3}}\Rightarrow {{T}_{n+1}}-{{T}_{n}}=21\] \[\Rightarrow \] \[^{n+1}{{C}_{3}}{{-}^{n}}{{C}_{3}}=21\] \[\Rightarrow \] \[^{n}{{C}_{2}}{{+}^{n}}{{C}_{3}}{{-}^{n}}{{C}_{3}}=21\] \[\Rightarrow \] \[^{n}{{C}_{2}}=21\] \[\Rightarrow \] \[\frac{n(n-1)}{2}=21\] \[\Rightarrow \] \[{{n}^{2}}-n-42=0\] \[\Rightarrow \] \[(n-7)(n+6)=0\] \[\therefore \] \[n=7\] As \[n\ge 1\]You need to login to perform this action.
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