If S1= 2 + 4 + ... + 2n and S2 = 1 + 3 + ... + (2n — 1), then S1: S2 = …..
We know that, Sn =
× (a + l)
Where a = first term of the AP.
And l = last term of the AP
For S1,
We have S1 =
× (2 + 2n)
⇒ S1 = n × (n + 1)
For S2,
We have S2 =
× (1 + 2n – 1)
⇒ S2 =
× (2n)
⇒ S2 = n2
Now
=
= ![]()
∴
= ![]()
∴ the correct option is (a)
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