If S_n denotes the sum of the first r terms of an A.P. Then, S_3n : (S_2n – S_n) is

Solution (By Examveda Team)

$$\eqalign{ & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right], \cr & {S_{2n}} = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]\,{\text{and}} \cr & {S_{3n}} = \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr & {\text{Now}}\,{S_{2n}} - {S_n} \cr} $$
$$ = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right] - $$     $$\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$ = \frac{n}{2}\left[ {4a + \left( {4n - 2} \right)d} \right] - $$     $$\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$\eqalign{ & = \frac{n}{2}\left[ {4a - 2a + \left( {4n - 2 - n + 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr & = \frac{1}{3}\left( {{S_{3n}}} \right) \cr & \therefore {S_{3n}}:\left( {{S_{2n}} - {S_n}} \right) \cr & = 3:1\,{\text{or}}\,\frac{3}{1} = 3 \cr} $$