Explanation

(c) Distance travelled by the body in $n^{th}$ second is given by
$S_{n} = u + \frac{a}{2}(2n - 1)$
$5 = u + \frac{a}{2}(2 \times 1 - 1)$
$5 = u + \frac{a}{2}\ \ \ \ \ \ldots..(i)$
$2 = u + \frac{a}{2}(2 \times 3 - 1)$
$2 = u + \frac{5}{2}a\ldots.(ii)$
Solving Eqs. (i) and (ii), we get
$a = - \frac{6}{4}{ms}^{- 2}$
$ie,\ body\ $is decelerating
$Given,\ \ mass = 4kg$
$\therefore F = m \times a = 4 \times \frac{6}{4} = 6\ N$