A body moving with a uniform acceleration crosses a distance of 15m in the 3rd second and 23m in the 5th second. The displacement in 10sec will be

$$\begin{gathered} given S_3=15m;let initial velocity=u;acceleration =a;substitute in \\ {S}_n=u+\frac{a}{2}(2n-1) \\ 15=U+\frac{1}{2}a(2\times 3-1)=U+\frac{5a}{2}\to \left(1\right) \\ given S_5=23m; substitute in \\ {S}_n=u+\frac{a}{2}(2n-1) \end{gathered}$$

$23=U+\frac{1}{2}a(2\times 5-1)=U+\frac{9a}{2}\to \left(2\right)$

Solving (1) and (2)

a   =   4m/$s^2$ , U = 5m/s

Displacement in 10 sec

$S=Ut+\frac{1}{2}a{t}^2$

$=5\times 10+\frac{1}{2}\times 4\times {10}^2$

S=   250 m=Displacement in 10 sec