Let Sn denote the sum of the first ‘n’ terms of an APS2n

Let Sn denote the sum of the first ‘n’ terms of an AP
S_2n = 3S_n Then, the ratio S_3n is equal to
S_n

Let a be the first term and d be the common difference.

Then, S_n =	n	[2a + (n - 1)d]
	2

S_2n =	2n	[2a + (2n - 1)d]
	2

and S_3n =	3n	[2a + (3n - 1)d]
	2

Given, S_2n = 3S_n

∴	2n	[2a + (2n - 1)d] = 3	n	[2a + (n - 1)d]
	2		2

⇒ 4a + (4n – 2)d = 6a + (3n – 3)d
⇒ d (4n – 2 – 3n + 3) = 2a

⇒ d =	2a
	n + 1

∴ S_n =	2an²
	n + 1

and S_3n =	12an²
	n + 1

∴	S_n	=	2an²	×	n + 1	=	1
	S_3n		n + 1		12an²		6

⇒	S_3n	= 6
	S_n