Subba Rao's incomes in various years form an arithmetic progression (A.P.) with an annual increment of Rs. \(200\). The salaries from 1995 onwards are: \(5000, 5200, 5400, ……\). Here, the first term is \(a = 5000\) and the common difference is \(d = 200\). If Subba Rao's salary reaches Rs. \(7000\) in the \(n^{th}\) year, we use the formula \(a_n = a + (n − 1) d\). Substituting the values: \(7000 = 5000 + (n − 1) 200\). This simplifies to \(200(n − 1) = 2000\), leading to \((n − 1) = 10\) and thus \(n = 11\).
Therefore, Subba Rao's salary will be Rs. \(7000\) in the \(11^{th}\) year.