In an AP , Sn = 2n , S2n = n Find S4n
S(n) = (n/2)(2a + (n-1)d) = 2n
divide by n and double
2a + (n-1)d = 4
S(2n) = (2n/2)(2a + (2n-1)d) = n
divide by n
2a + (2n-1)d = 1
subtract these two equations,
(n-1)d - (2n-1)d = 3
nd - d - 2nd + d = 3
-nd = 3
nd = -3
S(4n) = (4n/2)(2a + (4n-1)d)
= 2n(2a + 4nd - d)
= 2n(2a -12 - d)
= 4an - 24n - 2nd
Don't know if this gets you anywhere without knowing some more info