Here a = 3

a₂= 7

a₃ = 11

d= a3– a2 = a2– a1 = 4

S_n= (2a + (n–1) d)

406 = (6 + (n–1)4)

406 = (4n + 2)

2n² + n – 406 = 0

2n² + 29 n – 28 n – 406 = 0

n (2n + 29) – 14(2n + 29) = 0

(n – 14) (2n + 29) = 0

Number of terms cannot be negative and in fractions so n= 14