Given a = 1, an = 20 and Sn = 399.
We know that
an = a + (n - 1)d
∴ an = 1 + (n - 1) × d
⇒ 20 = 1 + (n - 1)d
⇒ (n - 1)d = 20 - 1
⇒ (n - 1)d = 19
⇒ d = n−119.
The formula for sum of A.P. is given by,
Sn=2n[2a+(n−1)d]∴Sn=2n[2×1+(n−1)×d]⇒399=2n[2+(n−1)×n−119]⇒399=2n[2+19]⇒2n×21=399⇒n=21399×2⇒n=21798⇒n=38.
Hence, Option 3 is the correct option.