OpenStudy (doc.brown):

An old man walks 2.7m out of his house, turns around, and walks back. Every day he walks 2.7m further than the day before. One day he finally fetches his mail from the mailbox 47m down the driveway and returns home. 1. How many days did it take to get the mail? 2. What was the total distance traveled?

OpenStudy (doc.brown):

@Isaiah.Feynman $$\frac{47}{2.7} = 17.4$$ It didn't take 0.4 of a day on his last trip. He made the last trip on the 18th day, if I draw it out.

OpenStudy (shamil98):

d = 47/2.7 17 and a half days ish.. so he travels 2.7 * 2 (the trip back counts too ) on the first day.. 5.4 meters on the first day.. thinking of a function/sequence to write this as..

OpenStudy (shamil98):

each day he goes + 5.4 meters more..

OpenStudy (shamil98):

So, each day he travels 5.4 meters more than the previous day.. er, still thinking of a sequence/function, sleepy..

OpenStudy (shamil98):

First day : 5.4 total Second day: 10.8 total common ratio is 2. Looking for the total distance, so i would suggest using the sum of a geometric series formula: $\huge \sum_{k=0}^{n-1} (ar^k) = a(\frac{ r^n - 1 }{ r-1 })$ not too sure tho

OpenStudy (doc.brown):

Can we do the sum of all the trips there and multiply it by 2 at the end, to get the trips back?

OpenStudy (shamil98):

or you could just let a = 5.4 instead of 2.8

OpenStudy (shamil98):

what class is this for? and is it a geometric series problem?

OpenStudy (doc.brown):

It's for real life, I'm trying to find out how much gravel I need on a path. This part of a very complicated formula is the only part giving me trouble.

OpenStudy (shamil98):

@Isaiah.Feynman what are your thoughts on this ?

OpenStudy (shamil98):

I think the geometric sum formula will give you the total distance.. you could also just count it by yourself but its tedious

OpenStudy (shamil98):

Oh wait my bad, it's an arithmetic series , each day its increased by 5.4 not multiplied by 2.. cause on day 3 you would have 16.2

OpenStudy (shamil98):

$\huge \frac{ n }{ 2 } (2a + (n-1)d)$ a = first term d = difference between terms n = the number of terms you need to add up

OpenStudy (shamil98):

The thing is do you want the exact distance? or do you want it rounded down to 17 days or up to 18 days? or just as a decimal?..

OpenStudy (doc.brown):

If I grunt it out $2(2.7+2(2.7)+3(2.7)+4(2.7)+5(2.7)$$+6(2.7)+7(2.7)+8(2.7)+9(2.7)+10(2.7)$$+11(2.7)+12(2.7)+13(2.7)+14(2.7)$$+15(2.7)+16(2.7)+17(2.7)+1.1)$$=828.4m$But I want the formula.

OpenStudy (shamil98):

Okay.

OpenStudy (shamil98):

$\huge \frac{ 18 }{ 2 }(2(5.4) + (17)5.4)$ $\huge 8.5 ( 10.8 + 91.8)$ $\huge= 923.4$

OpenStudy (shamil98):

typo that shud be a 9 * if n = 17... $\huge \frac{ 17 }{ 2 }(2(5.4) + 16(5.4))$ $\huge 8.5 (10.8 + 86.4)$ $\huge = 826.2$

OpenStudy (doc.brown):

still not 828.4m

OpenStudy (shamil98):

hm

OpenStudy (doc.brown):

I used @Isaiah.Feynman number from earlier$\frac{17.4}{2}(2(5.4)+(16.4)(5.4))$$8.7(10.8+88.56)$$=864.432$nope

OpenStudy (shamil98):

that's a geometric series and it's invalid.. each day he walks 2.7 meters more than the day before it's addition not multiplication: including the trip back: day 1 : 5.4 day 2 : 10.8 day 3 : 16.2

OpenStudy (agent0smith):

2(2.7+2(2.7)+3(2.7)+4(2.7)+5(2.7) +6(2.7)+7(2.7)+8(2.7)+9(2.7)+10(2.7) +11(2.7)+12(2.7)+13(2.7)+14(2.7) +15(2.7)+16(2.7)+17(2.7)+1.1) Where's the 18th day... Where did 1.1 come from?

OpenStudy (agent0smith):

On the 17th day he does not reach the mailbox, he only walks 45.9m. On the 18th day he walks 47m, stops and turns around. With the arithmetic sequence formula, you're adding a little bit too much distance (since it's 48.6 when n=18 so it predicts 1.6m too far). So if you subtract twice that extra distance, or 3.2m, you should have the correct distance.

OpenStudy (agent0smith):

Sum is$\Large \frac{ n }{ 2 } (2a + (n-1)d)$where n=18 (takes him 18 days), a=5.4 (dist walked on first day) d=5.4 (dist gained per day) Then subtract the extra 3.2m that the formula is adding, since he only walks 47m and turns around, not 48.6m $\Large \frac{ 18 }{ 2 } (2*5.4 + (18-1)*5.4) - 3.2 = 920.2m$

OpenStudy (shamil98):

i was close xD

OpenStudy (agent0smith):

This gigantic mess gives the same thing...I'm still confused why there was a 1.1 in it originally, though, and missing the 18th day. 2(2.7+2(2.7)+3(2.7)+4(2.7)+5(2.7) +6(2.7)+7(2.7)+8(2.7)+9(2.7)+10(2.7) +11(2.7)+12(2.7)+13(2.7)+14(2.7) +15(2.7)+16(2.7)+17(2.7)+18*(2.7))-3.2 = 920.2

OpenStudy (agent0smith):

^lol did you mean to say "Just not my night"? If so, it really isn't your night :P

OpenStudy (mathstudent55):

$$2(2.7) + 2(2.7 + 2.7) + 2(2.7 + 2.7 + 2.7) + ... + 2(17)(2.7) + 2(47)$$ $$=2(2.7) + 2(2)2.7 + 2(3)(2.7) + ... + 2(17)(2.7) + 2(47)$$ $$=5.4(1) + 5.4(2) + 5.4(3) + ... + 5.4(17) + 2(47)$$ $$=5.4(1 + 2 + ... + 17) + 94$$ $$= \large 5.4 \sum_{i=1}^{17} i + 94$$ Since $$\sum_{i = 1}^{n} i ~~=~~ \dfrac{n(n + 1)}{2}$$, then $$= 5.4 \cdot \dfrac{(17)(18)}{2} + 94$$ $$= 920.2$$ 920.2 m (assuming that on the 18th day he only walks 47 m each way.)

OpenStudy (doc.brown):

Ha! I see what I did, thanks @agent0smith. $17(2.7)+1.1) = 47$I was right to add it but then I forgot my 17(2.7) for the 17th trip.

OpenStudy (doc.brown):

Ok, I hate to do this to you. I'm going to change the question. A man walks length L, turns and walks back. On each trip he walks L further than the trip before. Finally he travels distance D and returns. What is the formula to find the total distance traveled?

OpenStudy (doc.brown):

?$2\left(L\sum^{roundDown(\dfrac{D}{L})}_{i=1}i+D\right)$?

OpenStudy (shamil98):

How do you write those huge brackets in latex? ;o

OpenStudy (doc.brown):

'\left(\right)'

OpenStudy (anonymous):

@ehuman

OpenStudy (doc.brown):

@agent0smith Let's say L=1.1 and D=2.5 $n=\frac{2.5}{1.1}=2.27=3$$a=2(1.1)=2.2$$\frac{n}{2}(2a+(n-1)d)+2(D)$$\frac{3}{2}(2(2.2)+(3-1)2.2)+2(2.5)$$\frac{3}{2}(4.4+4.4)+5$$=18.2$ But when I think about it I get|dw:1387787765675:dw|

OpenStudy (agent0smith):

Because you added a number on the end where you should be subtracting one. On the third day he only walks 2.5m, not the 3.3 the formula predicts. So you should be subtracting 2*0.8, not adding 2(2.5)... why were you adding that?

OpenStudy (doc.brown):

'Cause I'm a flippin idiot