A body moving with uniform acceleration describes 4 m in 3rd second and 12 m in the 5th second. The distance described in the next three seconds is

Let u be the initial velocity and o be the acceleration, the
${\mathrm{s}}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2\mathrm{n}-1)$
$\therefore {\mathrm{s}}_3=\mathrm{u}+\frac{\mathrm{a}}{2}(2\times 3-1)$
or 4-u+(5a/2)….(1)
Similarly ${\mathrm{s}}_5=\mathrm{u}+\frac{\mathrm{a}}{2}(2\times 5-1)$
or 12 = u+(9 a/2) ...(2)
Solving these equations, we get
u = -6m/s and a = 4m/s²
Distance travelled in 5 second
$=-6\times 5+\frac{1}{2}\times 4\times (5{)}^2=20\mathrm{m} \left(\because \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\right)$
Distance travelled in 8 second
$=-6\times 8+\frac{1}{2}\times 4\times (8{)}^2=80\mathrm{m}$
Distance travelled in next three second
=80 -20 =60m