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 is
A. 5

B. 10

C. 12

D. 14

Answer :

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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