please do this problem

let A 1 , A 2 , A 3 , A 4 ......A n be n Arithmetic means are instead between 20 and 80.
i.e., 20 , A 1 , A 2 , A 3 , A 4 ......A n 80.
∴ The Arithmetic Progression consist of ' n + 2 'terms.
let 'd' be the common difference.
The first term a = 20
The last term L = 80.
but the last term is T n + 2 = a + (n + 2 − 1)d
= a + (n + 1)d
T n + 2 = 80
a + (n + 1)d = 80
20 + (n + 1)d = 80
(n + 1)d = 60
d=60/n+1
Given that the ratio of the first mean to the last mean is 1:3
i.e.,A1/An=1/3
 
20 + d/20 + nd=1/3
(A n = T n + 1 = a + nd)
20 +60/n+1/20+n.60/n+1
3(n + 4) = (4n + 1)
3n + 12 = 4n + 1
n = 11.