AIME 2023 II · 第 1 题

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

$$\begin{aligned} 6a+15d &= 990, \\ 2a &= a+5d. \end{aligned}$$ The second equation implies that $a=5d.$ Substituting this into the first equation, we get

$$\begin{aligned} 6(5d)+15d &=990, \\ 45d &= 990 \\ d &= 22. \end{aligned}$$ 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

Video Solution by the Power of Logic(both #1 and #2)

https://youtu.be/VcEulZ3nvSI