Question

Which term of the A.P. $40,36,32,..$ is $0$. The sum of how many terms of the A.P is $0$.

Answer 1

Given,$a=40$

$d=36-40=-4$

$a_n=a+(n-1)d$

$0=40+(n-1)(-4)$

$40-4n+4=0$

$\therefore n=\frac{44}{4}=11$

$11th$ term is zero.

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$0=\frac{n}{2}\left[2\right(40)+(n-1\left)\right(-4\left)\right]$

$n[80-4n+4]=0$

$84n-4n^2=0$

$4n^2=84n$

$n=21$

sum of $21$ terms is zero.