Question

Is $0$ a term of the A.P. $31,28,25,...$? Justify.

Answer 1

$t_1=a=31$

$t_2=28$

$t_2-t_1=28-31=-3$ Since difference between consecutive terms is constant.

$d=-3$ $\therefore$ sequence is in A.P

$\therefore$ sequence is, $t_n=a+(n-1)d$

$0=31+(n-1)(-3)$

$0=31-3n+3$ $\therefore 34=3n$ $n=\frac{34}{3}=11.33$

which is not an integer

$\therefore$ $0$ is not a term of the following $A.P$