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Difference between revisions of "1999 AHSME Problems/Problem 20"

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== Problem ==
 
== Problem ==
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The sequence <math>a_{1},a_{2},a_{3},\ldots</math> statisfies <math>a_{1} = 19,a_{9} = 99</math>, and, for all <math>n\geq 3</math>, <math>a_{n}</math> is the arithmetic mean of the first <math>n - 1</math> terms.  Find <math>a_2</math>.
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The sequence <math>a_{1},a_{2},a_{3},\ldots</math> satisfies <math>a_{1} = 19,a_{9} = 99</math>, and, for all <math>n\geq 3</math>, <math>a_{n}</math> is the arithmetic mean of the first <math>n - 1</math> terms.  Find <math>a_2</math>.
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<math>\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 00:50, 17 December 2021

Problem

The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$, and, for all $n\geq 3$, $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$.

$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$

Solution

Let $m$ be the arithmetic mean of $a_1$ and $a_2$. We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$.

By definition, $a_3=m$.

Next, $a_4$ is the mean of $m-x$, $m+x$ and $m$, which is again $m$.

Realizing this, one can easily prove by induction that $\forall n\geq 3;~ a_n=m$.

It follows that $m=a_9=99$. From $19=a_1=m-x$ we get that $x=80$. And thus $a_2 = m+x = \boxed{(E) 179}$.

See also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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