A particle moves in a straight line with constant acceleration a. The displacements of particle from origin in times t1, t2 and t3 are s1, s2 and s3 respectively. If times are in AP with common difference d and displacements are in GP, then prove that a=s1-s3nd2.

Let us draw v-t graph of the given situation, area of which will give the displacement and slope the acceleration.s2-s1=xd+12yd....(i) s3-s2=xd+yd+12yd....(ii)Subtracting Eq. (i) from Eq. (ii), we haves3+s1-2s2=ydor s3+s1-2s1s3=yd  s2=s1s3Dividing by d2 both sides we have,s1-s32d2=yd= slope of v-t graph =a.