The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
Here a = 3
a2= 7
a3 = 11
d= a3– a2 = a2– a1 = 4
Sn= (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