A particle travels 10m in the first 5 sec and 10m in the next 3 sec. Assuming constant acceleration what is the distance travelled in the next 2 sec?

We know that $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

If the initial velocity is u and acceleration is a then distance in the first 5 seconds will be: 
$10=5\mathrm{u}+\frac{1}{2}\mathrm{a}\left(25\right)$     ...eq.(1)

Distance in 8 seconds will be:
 $\mathrm{s}=8\mathrm{u}+\frac{1}{2}\mathrm{a}\left(64\right)=8\mathrm{u}+32\mathrm{a}$

So distance between t=5 and t=8  will be:  $\mathrm{s}-10=8\mathrm{u}+32\mathrm{a}-10$
Distance between 5s and 8s is given as 10 therefore,
$10=8\mathrm{u}+32\mathrm{a}-10$
$8\mathrm{u}+32\mathrm{a}=20$              ...eq.(2)

Solving equation 1 and equation 2 we get $\mathrm{a}= \frac{1}{3}$ and $\mathrm{u}=\frac{35}{30}$

Now distance in 10 seconds is ${\mathrm{s}}_1=10\mathrm{u}+\frac{100\mathrm{a}}{2}=10\mathrm{u}+50\mathrm{a}$

So distance in the last two seconds will be:
${\mathrm{S}}_1-\mathrm{S}=10\mathrm{u}+50\mathrm{a}-(8\mathrm{u}+32\mathrm{a})$
$=2\mathrm{u}+18\mathrm{a}=\frac{30}{70}+\frac{18}{3}=\frac{25}{3}$$\approx 8.3 m$