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考题7-8 2015 AMC 10A

2018-08-06 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 7

How many terms are in the arithmetic sequence $13$ , $16$ , $19$ , $\dotsc$ , $70$ , $73$ ?

$考题7-8 2015 AMC 10A$

Solution

$73-13 = 60$ , so the amount of terms in the sequence $13$ , $16$ , $19$ , $\dotsc$ , $70$ , $73$ is the same as in the sequence $0$ , $3$ , $6$ , $\dotsc$ , $57$ , $60$ .

In this sequence, the terms are the multiples of $3$ going up to $60$ , and there are $20$ multiples of $3$ in $60$ .

However, one more must be added to include the first term. So, the answer is $\boxed{\textbf{(B)}\ 21}$ .

Solution 2

$考题7-8 2015 AMC 10A$ $\boxed{\textbf{(B)}}$ .

Solution 3

Using the formula for arithmetic sequence's nth term, we see that $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$

Solution 4

Minus each of the terms by 12 to make the the sequence 1,4,7,....,61. 61-1/3=20 20+1=21 $\boxed{\textbf{(B)}}$ .

Problem 8

Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ : $1$ ?