4th term,T4= a+3d = 11
It is given that,
8th term,T8 = a+ 7d= (2×T4 ) +5
So,
T8 = 2×11 +5 = 27
Therefore
a+3d = 11 ---->(1)
a+7d=27----->(2)
Solving
(2) - (1)
4d = 16
Therefore d = 16/4 =5
So, from(1)
a +3 ×4 = 11
So , a = 11-12 = -1
So AP is
-1,3,7,11,15,19,23,27 ......
Sumof 50 terms =
(n/2)×(2a +(n-1)×d)
=(50/2)×(2×-1 +(50-1)×4)
=25×(-2 +49×4) =25×194 = 4656