Given the 3rd and 5th terms of an A.P. are 6 and 10 respectively, we have: The formula for the \( n \)-th term of an A.P. is given by: \(T_n = a + (n-1)d\) where \( a \) is the first term and \( d \) is the common difference. For the 3rd term: \(T_3 = a + 2d = 6 \quad \text{(Equation 1)}\) For the 5th term: \(T_5 = a + 4d = 10 \quad \text{(Equation 2)}\) From these equations, we have: \(a + 2d = 6 \tag{1}\) \(a + 4d = 10 \tag{2}\) Subtract Equation 1 from Equation 2: \((a + 4d) - (a + 2d) = 10 - 6\) This simplifies to: \(2d = 4\) Thus, \(d = 2\) Substituting \( d = 2 \) into Equation 1: \(a + 2(2) = 6\) This simplifies to: \(a + 4 = 6\) Thus, \(a = 2\) The first term \( a \) is \( 2 \) and the common difference \( d \) is \( 2 \).