## 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