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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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== Solution ==
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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>
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== 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>.
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Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>.
 
Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>.
  
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It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \ (E) boxed{179}</math>.
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It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \boxed{(E) 179}</math>.
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== Solution 2==
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Let <math>a_1=a</math> and <math>a_2=b</math>. Then, <math>a_3=\frac{a+b}{2}</math>, <math>a_4=\frac{a+b+\frac{a+b}{2}}{3}=\frac{a+b}{2},</math> and so on.
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Since <math>a_3=a_4</math>, <math>a_n=a_3</math> for all <math>n\geq3.</math>
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Hence, <math>a_9=\frac{a_1+a_2}{2}=\frac{a+b}{2}=99, a+b=198.</math> We also know that <math>a_1=a=19.</math>
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Subtracting <math>a_1</math> from <math>198,</math> we get <math>b=a_2=\boxed{(E)  179}.</math>
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~Benedict T (countmath1)
  
 
== See also ==
 
== See also ==

Latest revision as of 13:00, 21 March 2023

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 $a_2 = m+x = \boxed{(E) 179}$.

Solution 2

Let $a_1=a$ and $a_2=b$. Then, $a_3=\frac{a+b}{2}$, $a_4=\frac{a+b+\frac{a+b}{2}}{3}=\frac{a+b}{2},$ and so on.

Since $a_3=a_4$, $a_n=a_3$ for all $n\geq3.$

Hence, $a_9=\frac{a_1+a_2}{2}=\frac{a+b}{2}=99, a+b=198.$ We also know that $a_1=a=19.$

Subtracting $a_1$ from $198,$ we get $b=a_2=\boxed{(E) 179}.$

~Benedict T (countmath1)

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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