The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
Answers
Answered by
6
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
Similar questions