Revision as of 23:48, 17 February 2023

Problem

The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees.

Solution 1

In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is $a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.$ We are given that $\begin{align*} 6a+15d &= 990, \\ 2a &= a+5d. \end{align*}$ The second equation implies that $a=5d.$ Substituting this into the first equation, we get $\begin{align*} 6(5d)+15d &=990, \\ 45d &= 990 \\ d &= 22. \end{align*}$ It follows that $a=110.$ Therefore, the greatest number of apples growing on any of the six trees is $a+5d=\boxed{220}.$

~MRENTHUSIASM

Solution 2

Let the terms in the sequence be defined as $a_1, a_2, ..., a_6.$

Since this is an arithmetic sequence, we have $a_1+a_6=a_2+a_5=a_3+a_4.$ So, $\sum_{i=1}^6 a_i=3(a_1+a_6)=990.$ Hence, $(a_1+a_6)=330.$ And, since we are given that $a_6=2a_1,$ we get $3a_1=330\implies a_1=110$ and $a_6=\boxed{220}.$

~Kiran

=Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=nNhfDCX5-bw

@@ Line 29: / Line 29: @@
 ~Kiran
-==Video Solution by SpreadTheMathLove=
+==Video Solution 1 by SpreadTheMathLove=
 https://www.youtube.com/watch?v=nNhfDCX5-bw

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Difference between revisions of "2023 AIME II Problems/Problem 1"

Revision as of 23:48, 17 February 2023

Contents

Problem

Solution 1

Solution 2

=Video Solution 1 by SpreadTheMathLove

See also