Solution :
Given, \(T_n\) = \(^nC_3\)
\(T_{n+1}\) = \(^{n+1}C_3\)
\(\therefore\) \(T_{n+1}\) – \(T_n\) = \(^{n+1}C_3\) – \(^{n}C_3\) = 10 [given]
\(\therefore\) \(^nC_2\) + \(^nC_3\) – \(^nC_3\) = 10
\(\implies\) \(^nC_2\) = 10
\(\therefore\) n = 5