A body covers 26, 28, 30, 32 meters in 10^th,11^th,12^th and 13^th seconds respectively. The body starts

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