Answer :

Given that: S_{n} denote the sum of first n terms

and S_{2n} = 3S_{n}

To find: S_{3n} : S_{n}

Now, we know that

⇒ S_{2n} = n[2a + (2n – 1)d]

As per the given condition of the question, we have

S_{2n} = 3S_{n}

⇒ 4an + 2nd(2n – 1) = 6an + 3nd(n – 1)

⇒ 2nd(2n – 1) – 3nd(n – 1) = 6an – 4an

⇒ 4n^{2}d – 2nd – 3n^{2}d + 3nd = 2an

⇒ nd + n^{2}d = 2an

⇒ nd(1 + n) = 2an

⇒ d(n + 1) = 2a …(i)

Now, we have to find S_{3n}:S_{n}

So,

[from (i)]

Hence, the correct option is **(b)**

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