smc

Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is $$S_1+S_2+S_3+...+S_n$$ $$=a_1+(a_1+a_2)+(a_1+a_2+a_3)+\dots +(a_1+a_2+\dots +a_n)$$ $$=n\cdot a_1+(n-1)\cdot a_2+\dots +a_n.$$ If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is $1000\ast 99=99,000,$ than the Cesáro total of the second sequence is $n\cdot a_1+99,000=100\cdot 1+99,000=99,100.$ Thus the Cesáro sum of the second sequence is $\frac{99,100}{100}=\boxed{991}.$