We know that, distance travelled in n th second sn=u+
a
2
(2n−1) Case 1 Given, n=3,s=12m So,12=u+
a
2
(6−1) ⇒12=u+2.5a .......(i) Case 2s=20m,n=5 So, 20=u+
a
2
(10−1) ⇒20=u+4.5a ........(ii) from Eqs. (i) and (ii), 8=a(4.5−25) ⇒8=2a ⇒a=4m∕s2 .......(iii) From Eqs. (i), ⇒12=u+25×4⇒u=2m∕s From first equation of motion, v=u+at v=2+4×10=42m∕s