bus holds 66 people. Starts out empty. Picks up 1 person at the 1st station. 2nd person at the 2nd station 3rd person at the 3rd station and so on. No one gets off. After how many stations will fill up the bus Show your work- explain your thinking?
2 answers
Wouldn't the last person get on at the 66th station?
the way you worded the question it simply becomes
1+1+1+1+... = 66
clearly 66 ones are needed, so after 66 stations the bus is full
if you meant 1st station one person gets one, on the 2nd station two people get on, on the third 3 people get on, etc
then it becomes a sequence/series type of question and you would have
1+2+3+4+... = 66
here a=1
d=1 and n is unknown but S(n) = 66
S(n) = n/2[2a + d(n-1)]
66 = n/2[2 + 1(n-1)]
132 = n[1+n]
n^2 + n - 132 = 0
(n+12)(n-11) = 0
n = -12 or n = 11
we need the positive integer solution which is 11
test: S(11) = 11/2[2 + 10] = 66
1+1+1+1+... = 66
clearly 66 ones are needed, so after 66 stations the bus is full
if you meant 1st station one person gets one, on the 2nd station two people get on, on the third 3 people get on, etc
then it becomes a sequence/series type of question and you would have
1+2+3+4+... = 66
here a=1
d=1 and n is unknown but S(n) = 66
S(n) = n/2[2a + d(n-1)]
66 = n/2[2 + 1(n-1)]
132 = n[1+n]
n^2 + n - 132 = 0
(n+12)(n-11) = 0
n = -12 or n = 11
we need the positive integer solution which is 11
test: S(11) = 11/2[2 + 10] = 66