Q. 195.0( 1 Vote )

# The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 isA. 5B. 10C. 12D. 14

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

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