Difference between revisions of "1984 AIME Problems/Problem 1"

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== Problem ==
 
== Problem ==
Find the value of <math>\displaystyle a_2+a_4+a_6+a_8+\ldots+a_{98}</math> if <math>\displaystyle a_1</math>, <math>\displaystyle a_2</math>, <math>\displaystyle a_3\ldots</math> is an arithmetic progression with common difference 1, and <math>\displaystyle a_1+a_2+a_3+\ldots+a_{98}=137</math>.
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Find the value of <math>\displaystyle a_2+a_4+a_6+a_8+\ldots+a_{98}</math> if <math>\displaystyle a_1</math>, <math>\displaystyle a_2</math>, <math>\displaystyle a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>\displaystyle a_1+a_2+a_3+\ldots+a_{98}=137</math>.
  
 
== Solution ==
 
== Solution ==
{{solution}}
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One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to find the value of <math>a_1</math>, then use that to calculate <math>a_2</math> and sum another arithmetic series to get our answer.
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A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>.  We can substitute this into our given equation to get <math>(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137</math>.  The left-hand side of this equation is simply <math>2(a_2 + a_4 + \ldots + a_{98}) - 49</math>, so our desired value is <math>\frac{137 + 49}{2} = 093</math>.
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== See also ==
 
== See also ==
 
* [[1984 AIME Problems/Problem 2 | Next problem]]
 
* [[1984 AIME Problems/Problem 2 | Next problem]]
 
* [[1984 AIME Problems]]
 
* [[1984 AIME Problems]]

Revision as of 11:05, 24 January 2007

Problem

Find the value of $\displaystyle a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $\displaystyle a_1$, $\displaystyle a_2$, $\displaystyle a_3\ldots$ is an arithmetic progression with common difference 1, and $\displaystyle a_1+a_2+a_3+\ldots+a_{98}=137$.

Solution

One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$, then use that to calculate $a_2$ and sum another arithmetic series to get our answer.

A somewhat quicker method is to do the following: for each $n \geq 1$, we have $a_{2n - 1} = a_{2n} - 1$. We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$. The left-hand side of this equation is simply $2(a_2 + a_4 + \ldots + a_{98}) - 49$, so our desired value is $\frac{137 + 49}{2} = 093$.

See also