Q6 of 67 Page 74

If the sum of the first n terms of the arithmetic progression is Sn and S2n = 3Sn, then S3n : Sn will be :

Since the sum of n terms is
Sn =


Where,


a = First term of AP
d = Common difference of AP
and no of terms is ‘n’



⇒ S2n = 3Sn
⇒ 2[2a + (2n - 1)d] = 3[2a + (n - 1)d]


⇒ 4a + (2n - 1)2d] = 6a + (n - 1)3d
⇒ 4nd – 2d = 2a + 3nd – 3d
⇒ nd + d = 2a
⇒ (n + 1)d = 2a


S3n:Sn =
=
=
=
=
= 6
S3n : Sn = 6:1

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