A body moving with a uniform acceleration travels 4 m in the 3rd second and 12 m in the  5^th second. The distance travelled by the body in the next five seconds is

Distance travelled in 3^rd second
S₃ = 4m
Distance travelled in 5^th second
$\begin{array}{l}{\mathrm{s}}_5=12\mathrm{m}\\ 4=\mathrm{u}+\left(3-\frac{1}{2}\right)\mathrm{a}\\ 4=\mathrm{u}+\frac{5}{2}\mathrm{a}\end{array}$
$\begin{array}{l}8=2\mathrm{u}+5\mathrm{a}\to \left(1\right)\\ 12=\mathrm{u}+\left(5-\frac{1}{2}\right)\mathrm{a}\\ 24=2\mathrm{u}+9\mathrm{a}\to \left(2\right)\end{array}$
$\begin{array}{c}24=2\mathrm{u}+9\mathrm{a}\\ 8=2\mathrm{u}+5\mathrm{a}\\ - - -\\ \overline{16 =4}\mathrm{a}\end{array}$
From (1) and (2)
$\begin{array}{l}\mathrm{a}=\frac{16}{4}\\ \mathrm{a}=4{\mathrm{ms}}^{-2}\end{array}$
Substitute a value in eq (1)
$\begin{array}{l}8=2\mathrm{u}+5\times 4\\ 8=2\mathrm{u}+20\\ -2\mathrm{u}=20-8\\ -2\mathrm{u}=12\\ \mathrm{u}=-6{\mathrm{ms}}^{-1}\end{array}$
Distance travelled by the body in 5 s is
$\begin{array}{l}{\mathrm{s}}_1=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ {\mathrm{s}}_1=-6\times 5+\frac{1}{2}\times 4\times 5^2\\ {\mathrm{s}}_1=-30+2\times 25\\ {\mathrm{s}}_1=20\mathrm{m}\end{array}$
Distance travelled by the body in 10 s is

$$\begin{gathered} s_2=-6\times 10+\frac{1}{2}\times 4\times {10}^2 \\ {s}_2=-60+2\times 100 \\ {s}_2=-60+200 \\ {s}_2=140 m \end{gathered}$$

Distance travelled by the body in next 5 s is
 $\begin{array}{l}\mathrm{s}={\mathrm{s}}_2-{\mathrm{s}}_1\\ \mathrm{s}=140-20\\ \mathrm{s}=120\mathrm{m}\end{array}$