Difference between revisions of "1992 AHSME Problems/Problem 21"

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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
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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 <math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:06, 11 January 2018

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 $\fbox{A}$.

See also

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