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

m (Problem)
(Solution)
Line 4: Line 4:
 
<math>\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179</math>
 
<math>\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179</math>
  
== Solution ==
+
== Solution 1==
  
 
Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>.
 
Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>.

Revision as of 09:15, 17 April 2022

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 1

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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png