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

given S3=15m;let initial velocity=u;acceleration =a;substitute in Sn=u+a2(2n-1) 15  =   U+12 a (2×3−1)  =   U+5a2        →    (1) given S5=23m; substitute in Sn=u+a2(2n-1)23  =   U+12 a (2×5−1)  =   U+9a2        →    (2)Solving (1) and (2)a   =   4m/s2 , U = 5m/sDisplacement in 10 secS  =   Ut+12 a  t2=   5×10+12×4×102S=   250 m=Displacement in 10 sec