Question

The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is

Answer 1

Answer:

The AP is 3,7,11,15

Explanation:

then, ATP

a=3, d=7-3=4 and Sum Sn =406

therefore

sn = n/2[2a + ( n-1)×d]

=》406= n/2[2×3(n-1)×4]

=》812= n(6+4n-4)

=》812=n(4n+2)

=》4n^2+2n-812=0

=》2n^2+n-406=0

=》2n^2+29n-28n-406=0

=》n(2n+29)-14(n+29)=0

=》(2n+29)(n-14)=0

n=14 or n= -29/2

But n= -29/2 is not possible