Explanation

The distance covered in $n^{th} \text{second}$ is $S_{n} = u + \frac{1}{2}(2n - 1)a$ where u is initial velocity &a is acceleration then $26 = u + 19a/2$ $28 = u + 21a/2$ $30 = u + 23a/2$ $32 = u + 25a/2$ From eqs. (i) and (ii) we get $u = 7 m/\text{sec}$, $a = 2 m/\text{sec}^{2}$ $\therefore$ The body starts with initial velocity $u = 7 m/\text{sec}$ and moves with uniform acceleration $a = 2 m/\text{sec}^{2}$