$\therefore {S_1} = \frac{1}{2}a{(10)^2} = 50a$ .....(i)
$As\;\;v = u + at$
$\therefore $ velocity acquired by particle in $10 \,sec$ $v = a \times 10$
For next $10\, sec$ ,
${S_2} = (10a) \times 10 + \frac{1}{2}(a) \times {(10)^2}$
${S_2} = $ $150a$ .....(ii)
From (i) and (ii)
${S_1} = {S_2}/3$