(2n−1)a So distance travelled in tth&(t+1)th second are St=u+
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(2t−1)a St+1=u+
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(2t+1)a As per question, St+St+1=100=2(u+at)⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1) Now from first equation of motion the velocity of particle after time t , if it moves with an acceleration a is v=u+at⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(2) where u is initial velocity So from equation (1) and (2) , we get v=50cm∕s