It is known that the tolerance of {an} is 1, and a1 + A2 + +A98 + A99 = 99, then A3 + A6 + A9 + +a96+a99=? 55. I still don't understand.

It is known that the tolerance of {an} is 1, and a1 + A2 + +A98 + A99 = 99, then A3 + A6 + A9 + +a96+a99=? 55. I still don't understand.


The general formula of arithmetic sequence is
an=a1+(n-1)d
a1+a2+...+a99
=a1+a1+d+a1+2d+...+a1+98d
=99a1+d+2d+...+98d
=99a1+98(98+1)/2d
=99a1+4851=99
a1=(99-4851)/99=-48
a3+a6+a9+...+a99
=33a1+2d+5d+8d+...+98d
=-33*48+1650
=66