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November 26, 2014

November 26, 2014

Posted by **decker** on Wednesday, January 21, 2009 at 10:00am.

- question -
**bobpursley**, Wednesday, January 21, 2009 at 10:06amWouldn't the last person get on at the 66th station?

- question -
**Reiny**, Wednesday, January 21, 2009 at 10:15amthe 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

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