Question

a bus moves away from a bus stop in a straight line. it moves 10 meters by the end of the 1st second, 20 meters by the end of the 2nd second, and 30 meters by the end of the 3rd second. how many meters away is the bus from the bus stop after 10 seconds?
a. 100
c. 120
b. 110
d. 190
please select the best answer from the choices provided

Explanation:

Step1: Identify the sequence pattern

The distances form an arithmetic - sequence with first term $a_1 = 10$ and common difference $d=10$.

Step2: Use the sum formula for an arithmetic sequence

The sum formula of the first $n$ terms of an arithmetic sequence is $S_n=\frac{n(a_1 + a_n)}{2}$, where $a_n=a_1+(n - 1)d$. First, find $a_{10}$: $a_{10}=a_1+(10 - 1)d=10+(9)\times10=100$. Then find $S_{10}$: $S_{10}=\frac{10\times(10 + 100)}{2}$.

Step3: Calculate the sum

$S_{10}=\frac{10\times110}{2}=550$. But we can also use the formula $S_n=na_1+\frac{n(n - 1)}{2}d$. Substituting $n = 10$, $a_1=10$ and $d = 10$: $S_{10}=10\times10+\frac{10\times9}{2}\times10=100 + 450=550$. Another way is to directly sum: $10+20+30+\cdots+100=10\times(1 + 2+\cdots+10)$. Since the sum of the first $n$ positive integers is $\frac{n(n + 1)}{2}$, when $n = 10$, $\frac{10\times(10 + 1)}{2}=55$. And $10\times55 = 550$. However, if we assume the question means the distance in the 10th - second only:
The sequence of distances in each second is $a_n = 10n$. When $n = 10$, $a_{10}=10\times10=100$.

Answer:

A. 100