Question

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?

Answer 1

@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.

Answer 2

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..

Answer 3

each day he goes + 5.4 meters more..

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

or you could just let a = 5.4
instead of 2.8

Answer 8

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

Answer 9

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.

Answer 10

@Isaiah.Feynman what are your thoughts on this ?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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?..

Answer 15

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.

Answer 16

Okay.

Answer 17

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

Answer 18

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

Answer 19

still not 828.4m

Answer 20

hm

Answer 21

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

Answer 22

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

Answer 23

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?

Answer 24

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.

Answer 25

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
\[\Large \frac{ 18 }{ 2 } (2*5.4 + (18-1)*5.4) - 3.2 = 920.2m\]

Answer 26

i was close xD

Answer 27

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

Answer 28

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

Answer 29

\(2(2.7) + 2(2.7 + 2.7) + 2(2.7 + 2.7 + 2.7) + ... + 2(17)(2.7) + 2(47) \)

\(=2(2.7) + 2(2)2.7 + 2(3)(2.7) + ... + 2(17)(2.7) + 2(47) \)

\(=5.4(1) + 5.4(2) + 5.4(3) + ... + 5.4(17) + 2(47) \)

\(=5.4(1 + 2 + ... + 17) + 94\)

\(= \large 5.4 \sum_{i=1}^{17} i + 94\)

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.)

Answer 30

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.

Answer 31

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?

Answer 32

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

Answer 33

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

Answer 34

'\left(\right)'

Answer 35

@ehuman

Answer 36

@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

Answer 37

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?

Answer 38

'Cause I'm a flippin idiot