1992 AHSME Problems/Problem 21

Problem

For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the Cesáro sum of A is defined to be $\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$ . If the Cesáro sum of the 99-term sequence $(a_1,...,a_{99})$ is 1000, what is the Cesáro sum of the 100-term sequence $(1,a_1,...,a_{99})$ ?

$\text{(A) } 991\quad \text{(B) } 999\quad \text{(C) } 1000\quad \text{(D) } 1001\quad \text{(E) } 1009$

Solution

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]) ... = n*a[1] + (n - 1)*a[2] + ... + 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 * 99 = 99,000, than the Cesáro total of the second sequence is n*a[1] + 99,000 = 100 * 1 + 99,000 = 99,100. Thus the Cesáro sum of the second sequence is 99,100/100 = 991 $\fbox{A}$ .

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1992 AHSME Problems/Problem 21

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