1992 AMC 12 Problem 21

Question

For a finite sequence

A=(a_1,a_2,\ldots,a_n)

of numbers, the Cesàro sum of

A

is defined to be

\frac{S_1+S_2+\cdots+S_n}{n},

where

\begin{aligned} S_k&=a_1+a_2+a_3+\cdots+a_k\\ &\qquad(1\le k\le n). \end{aligned}

If the Cesàro sum of the

99

-term sequence

(a_1,a_2,\ldots,a_{99})

is

1000,

what is the Cesàro sum of the

100

-term sequence

(1,a_1,a_2,\ldots,a_{99})?

991

999

1000

1001

1009

Convert the first Cesàro sum into the value of

S_1+\cdots+S_{99}

Each new partial sum is

1

plus an old partial sum, with one initial partial sum equal to

1

The given condition says

S_1+\cdots+S_{99}=99{,}000.

For the new sequence, the partial sums are

1,1+S_1,\ldots,1+S_{99}.

Their sum is

100+(S_1+\cdots+S_{99})=99{,}100.

Dividing by

100

gives the new Cesàro sum

991.

Thus the correct answer is A.

Problem 21 in other years