Question

A1=12−12,a2=23−13,a3=34−14,a4=45−15,a5=56−16. a1=12−12,a2=23−13,a3=34−14,a4=45−15,a5=56−16.

Answer 1

$a_1=12-12=(1\times12-0)-(12+0)$
$a_2=23-13=(2\times12-1)-(12+1)$
$a_3=34-14=(3\times12-2)-(12+2)$
$a_4=45-15=(4\times12-3)-(12+3)$
$a_5=56-16=(5\times12-4)-(12+4)$

Clearly, the general for the $n$-th term is

$a_n=(12n-(n-1))-(12+n-1)=10n-10$

(Aside: given the simple pattern, it's curious why the terms would be given the way they are. It's easier to divine $10n-10$ from $a_1=0,a_2=10,a_3=20,\ldots$, but that's just my opinion.)

The sequence does not converge. As $n\to\infty$, so does $a_n\to\infty$.