If the distance covered by a body in nth second is given (4

+6n)Then find the accelaration and initial speed

S_nth = u + a(n-1/2)

       = u -a/2 + an

S_nt = 4 + 6n             (given)

comparing both equations

 u-a/2 = 4 ,  a = 6m/s²

 u = 4+a/2 = 7m/s

approve my ans if u like it

distance covered in nth second

S=4+6n .......1

also

S=u+a(2n-1)/2 .........2

differentiating equation 1 with respect to n

dS/dn=0+6

dS/dn=6 ......3

differentiating equation second with respect to n

dS/dn=0+a

dS/dn=a

putting value of dS/dn

a=6m/s^2

put the value of a in equation 2

and equate equation 1 and 2

4+6n=u+6(2n-1)/2

4+6n=6n-3+u

u=7m/s

Sn= u+a/2(2n-1)Also, Sn=u+an-a/2 or u-a/2+an.......1) and, Sn =4+6n..........2) comparing 1) and 2) we get :- u-a/2=4 and, an=6n. As, an=6n ;a=6m/s sq. Also, u-a/2=4i.e. u- 6/2=4 so, u-3=4 and u=4+3 or u= 7m/s.