1992 AHSME Problems/Problem 21

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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 1

Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is \[S_1 + S_2 + S_3 + ... + S_n\] \[= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)\] \[= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.\] 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 $n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.$ Thus the Cesáro sum of the second sequence is $\frac{99,100}{100} = \boxed{991, A}\, .$

Solution 2

We know that the Cesáro sum of the first $99$ numbers is

\[\frac{a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}}{99}=100.\]

Multiplying by $99$, we see that

\[a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}=100\cdot{99}.\]

Now, we append $1$ to the list of $a_n,$ so our new sum becomes

\[\frac{1+1+a_1+1+a_1+a_2+...+a_{99}}{100} \rightarrow\] \[\frac{100+a_1+a_1+a_2+...+a_{99}}{100}.\]

We've estabished that $a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}=100\cdot{99}$, so

\[\frac{100+a_1+a_1+a_2+...+a_{99}}{100}=\frac{100+99\cdot{100}}{100}.\]

Dividing everything by $100$, the sum of the sought sequence is $1+990=991 \rightarrow \boxed{A}$

~Benedict T (countmath1)

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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