AMC10 2021 A

AMC10 2021 A · Q4

AMC10 2021 A · Q4. It mainly tests Arithmetic sequences basics, Arithmetic misc.

A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$-second time interval, it travels $7$ inches more than during the previous $1$-second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?

一辆车滚下山坡，第一秒行进$5$英寸，并加速使得每连续的$1$秒时间间隔比前一个间隔多行进$7$英寸。车用了$30$秒到达山底。它总共行进了多少英寸？

(A) 215 215

(B) 360 360

(D) 3195 3195

(E) 3242 3242

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since \[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\] we seek the sum \[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\] in which there are $30$ terms. The last term is $5+7\cdot(30-1)=208.$ Therefore, the requested sum is \[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{\textbf{(D)} ~3195}.\] Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms: \[\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.\]

由于\[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\]我们求和\[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\]共有$30$项。末项为$5+7\cdot(30-1)=208$。因此，所求和为\[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{\textbf{(D)} ~3195}\]。回忆等差数列求和公式：首项和末项平均值乘以项数：\[\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.\]

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