Arithmetic progression

[Saumyojit]

A truck spends the day depositing truckloads of gravel from a quarry at equally spaced intervals along a straight road. The first load is deposited 20km from the quarry, the last is 10km further along the road. If the truck travels 550 km during the day, how many trips does it make, and how far apart are the deposits?

I got it the first deposition point is 20km away from quarry and the last is 30km from quarry . The truck goes each time to the quarry back and forth to fill up the next load at the next depositon point. Here in this case each Deposition Point denotes each term of AP.

Q1:What is meant by trip? a trip is a journey . What does one journey consists of in this case? Is the first trip/Journey will be considered when truck covers distance from (i)Q to d1 or (ii)Q to d1 and d1 to Q . Which of these 2 cases will be considered as first trip and why?

Q2:Wht the questioneer means by asking to find the no of trips ..it seems that he is wanting to find the no of deposition points . But how depostion points and no of trips are the same thing in what sense?. My intution suggests that one trip means completing the whole work of that deposition point (from picking up the load " from Quarry to depositing at each of Di points" and This part i am unsure---> "Coming Back to quarry" . this is why i asked Q1

I know the Formula which will be used to find n --> 550=n/2(first term+last term)

First clear this much . I have also a picture to share . i have done something by my own ; by knowing the n and common diffrence from the answers; tried to back calculate but i will upload it later.
@Dr.Peterson

 

[Dr.Peterson]

A truck spends the day depositing truckloads of gravel from a quarry at equally spaced intervals along a straight road. The first load is deposited 20km from the quarry, the last is 10km further along the road. If the truck travels 550 km during the day, how many trips does it make, and how far apart are the deposits?

I got it the first deposition point is 20km away from quarry and the last is 30km from quarry . The truck goes each time to the quarry back and forth to fill up the next load at the next deposition point. Here in this case each Deposition Point denotes each term of AP.

Q1:What is meant by trip? a trip is a journey . What does one journey consists of in this case? Is the first trip/Journey will be considered when truck covers distance from (i)Q to d1 or (ii)Q to d1 and d1 to Q . Which of these 2 cases will be considered as first trip and why?

This is as ambiguous to me as to you, but I would take a "trip" as two-way, from Q to Q, mostly because that is the same as the number of deposits, and there is no need to count one-way trips. Likewise, I assume the truck starts and ends the day at the quarry.

Q2:What the questioner means by asking to find the no of trips ..it seems that he is wanting to find the no of deposition points . But how deposition points and no of trips are the same thing in what sense?. My intuition suggests that one trip means completing the whole work of that deposition point (from picking up the load " from Quarry to depositing at each of Di points" and This part i am unsure---> "Coming Back to quarry" . this is why i asked Q1

I agree; one trip is (probably) from the quarry to a deposition point and back. But it doesn't make much difference apart from the number you use in your final answer.

And one of the first things I would do in trying to understand the meaning of the problem is to draw a picture; in fact, I did that, though only in my mind. What you describe is an appropriate method. We can talk about alternative methods when you finish.

 

[JeffM]

On the assumption that a trip means a round trip and the truck returns to the quarry at the end of the day, do the following equations make sense? If so why? Can you solve them?

[MATH]\text {Let } d = \text {distance between dump sites.}[/MATH]
[MATH]\text {Let } r = \text {number of round trips.}[/MATH]
[MATH]550 = 2 * \left ( \sum_{j=1}^r 20 + (j -1)d \right ) \text { and } (r - 1)d = 30 - 20 = 10.[/MATH]

 

[Dr.Peterson]

I know the Formula which will be used to find n --> 550=n/2(first term+last term)

This is a particularly efficient approach, akin to the method I used to solve the problem mentally. There will be a little twist at the end where one could get confused, but otherwise it takes less thought, and almost no algebra.

 

[Saumyojit]

there is no need to count one-way trips.

Even if i take one way trip whats the loss ? Q->d1, Q->d2,Q->d3,..STILL that is the same as the number of deposits .

Likewise, I assume the truck starts and ends the day at the quarry

imagine after the truck put the last load on the nth deposition point Then it has covered 550 km at that moment so it does not matter even if the truck returns to the quarry from the LAST deposition point. SO WHY do u say "ends the day at the quarry" --> it does not hold any significance

meaning of the problem is to draw a picture

i did that 2 days back only. But if i use the formula =>550=n/2(first term+last term)=AS each Deposition Point denotes each term of AP so n/2( Deposition Point1 +Deposition Point n th)

Now deposition points will be of some value and it unit is km. Deposition point 1 will be 20 km right? Deposition point (nth) will be 30 km
if this is true that means Value of each deposition point= Distance of it from the quarry . AM i right?

There will be a little twist at the end where one could get confused

are u saying we will divide 550 the sum by 2

 

[JeffM]

Even if i take one way trip whats the loss ? Q->d1, Q->d2,Q->d3,..STILL that is the same as the number of deposits .
imagine after the truck put the last load on the nth deposition point Then it has covered 550 km at that moment so it does not matter even if the truck returns to the quarry from the LAST deposition point. SO WHY do u say "ends the day at the quarry" --> it does not hold any significance

WHAT?

If the truck makes a trip back to the quarry after the last deposit, that is a trip of 30 km and is included in the 550. If it doesn't return to the quarry, that 30 km is not included in the 550.

If the problem does not specify that explicitly or implicitly, the problem cannot be solved. To put it a different way, is the last trip 30 km or 60 km? Why would you think that particular information is irrelevant?

are u saying we will divide 550 the sum by 2

If the last trip is not a round trip, then the drives to the deposit sites total (550 - 30)/2 = 260. If the last trip is a round trip, then the drives to deposit sites total 550/2 = 275. Of course, you don't think that is relevant.

EDIT In your first post, you asked what is a "trip." That is an excellent question, but it is not a mathematical one. If the problem is given in English and you quote it exactly and completely, we might be able to answer that question

 

[Dr.Peterson]

Even if i take one way trip whats the loss ? Q->d1, Q->d2,Q->d3,..STILL that is the same as the number of deposits .
imagine after the truck put the last load on the nth deposition point Then it has covered 550 km at that moment so it does not matter even if the truck returns to the quarry from the LAST deposition point. SO WHY do u say "ends the day at the quarry" --> it does not hold any significance
i did that 2 days back only. But if i use the formula =>550=n/2(first term+last term)=AS each Deposition Point denotes each term of AP so n/2( Deposition Point1 +Deposition Point n th)

Now deposition points will be of some value and it unit is km. Deposition point 1 will be 20 km right? Deposition point (nth) will be 30 km
if this is true that means Value of each deposition point= Distance of it from the quarry . AM i right?

are u saying we will divide 550 the sum by 2

Just show us your work, and we can discuss these issues. You should be able to see why these issues all matter; if not, we'll be able to point out why within your own work.

 

[Saumyojit]

@Dr.Peterson @JeffM PLEASE READ EACH SECTION. Its A humble request. I have taken time to write this post thinking carefully what should i ask or not.

If it doesn't return to the quarry, that 30 km is not included in the 550.

I understood that 550 km includes the journey from last depostion point to quarry only if i interpret trip as round.

is the last trip 30 km or 60 km?

If i consider the AP is happening in forward direction only then i can take a trip as a one-way trip but not a round trip.
THEN

CASE 1: THE LAST TRIP IS 30
My equation will be --> 550=n/2(20+30)
But i have to divide 550 by 2 as i am considering sum as a one way trip i.e 275km
275=n/2 * 50 --> n=11 (UNDERSTOOD THIS PART)
CASE 2: If ROUND TRIP
LAST TRIP IS 60
My equation will be --> 550=n/2 (40+60) -->n=11
HUGE DOUBT: THIS Equation i somehow created or got everything right in place in order to get n=11 but dont get the logic of how the formula works inside . In case 1 i have understood the equation and the working of formula (refernce to video) but in this case i am having problemn. NOTE: I know the reason why first trip is 40(20 forth 20 back) and last is 60 . I got the logic .but the working of formula in 2nd case not getting it.

I have a gut feeling u will not get me with what actually i am asking in this para named "HUGE DOUBT"

if it happens just once see this video "

" I will later continue with this discussion tomorrow morning .

Just show us your work

   20km       1km     1km                  

Quarry-----D1------D2--------D3...................Dn

         (20km)   (21km)    (22km)

If i consider one way trip then i can solve the sum in this way
550=n/2(20+30)
But i have to divide 550 by 2 as i am considering sum as a one way trip i.e 275km
275=n/2 * 50 --> n=11

Value of each deposition point= Distance of it from the quarry . AM i right?

Another thing i found out that Value of each deposition point (Each one way trip)= Distance of that point from the quarry
Just say right or wrong.

 

[Dr.Peterson]

You are getting the math right, but getting too tied up in the English, which we have already said is not quite clear. I don't think you should waste your time worrying about two interpretations. That gets in the way of seeing that you understand the math.

I take the problem,

If the truck travels 550 km during the day, how many trips does it make, and how far apart are the deposits?​

as meaning he makes one round trip for each deposit, the first 40 km and the last 60 km, and the total of those round trips is 550 km. Just as you say, this implies that n/2 (40+60) = 550, so that 50n = 550, and n = 11. Therefore there are 11 round trips, and the distance between deposits is (as you never actually stated, but implied in your picture near the end) 1 km, because 11 deposits means 10 spaces between deposits that total 10 km. (This is the point at which I mentioned a "twist at the end", as some people, not thinking deeply enough, would just divide 10 km by 11.)

So you have done well.

The way I personally did it was to think of the formula as, the sum of an arithmetic series is the number of terms times the average of the first and last terms. That is, 550 equals n times (40+60)/2, which is 50. So n is 550/50 = 11.

I really don't care about the "one-way trip" interpretation, because I don't think it is likely, and I've told you that. But if I took "trips" in the question as one-way trips, I would do exactly the same work, and simply say at the end that there were 22 one-way trips, since each round trip consists of two one-way trips. As I said in post #2, "it doesn't make much difference apart from the number you use in your final answer".

But in this interpretation, your equation 550=n/2(20+30) is wrong, because the sum of the one-way trips would only be half of the total distance. So your correction is right. But then you are taking n to mean only half the number of one-way trips, so your answer would be wrong! I suppose you are thinking of n as the number of deposits, not the number of trips. You never said what n meant.

Now, I have no idea why you think you understand the formula in the one case, but not in the other, because it's the same formula! You just applied it to different numbers. Maybe your problem is that you never wrote down what you were summing. In your case 1, you were summing this: 20 + ... + 30 = 275. The formula is the number of terms, n, times the sum of the first and last, 20+30, divided by 2. In case 2 (and what I did), we are summing 40 + ... + 60 = 550, and the formula works the same way.

I think by trying to do your "one-way" interpretation, but not being clear on what that means (e.g. what n stands for), you have confused yourself.

 

[JeffM]

You do not need to doubt.

The video says that the sum of n expressions with a common difference of d is

[MATH]\dfrac{n}{2} * (F + L),[/MATH] where F is the first term and L is the last term.

If we think about a trip as a round trip, the first trip is 40 and the last trip is 60 and the total distance travelled is 550. The formula says

[MATH]550 = \dfrac{n}{2} * (40 + 60) \implies n = 11.[/MATH]
If we think in terms of one way trips, we get that the first trip is 20 and the last is 30. But then the total distance driven one way is 550/2. So the formula says

[MATH]\dfrac{550}{2} = \dfrac{n}{2} * (20 + 30) \implies n = 11.[/MATH]
Is this a coincidence? Not at all. The second formula is exactly equivalent to the first.

[MATH]\dfrac{550}{2} = \dfrac{n}{2} * (20 + 30) \implies 2 * \dfrac{550}{2} = 2 * \dfrac{n}{2} * (20 + 30) \implies \\ 550 = \dfrac{n}{2} * 2(20 + 30) \implies 550 = \dfrac{n}{2} * (40 + 60).[/MATH]By the way: great video. Thanks.

Now that you have n = 11, be careful!

 

[Saumyojit]

This is the point at which I mentioned a "twist at the end", as some people, not thinking deeply enough, would just divide 10 km by 11

see after i find 'n' =11 I know that i have 2 deposit points given in question already i.e first and last so i need to place the rest 9 in between them . Thats how cd=1km i found out .

11 deposits means 10 spaces between deposits

Just out of curiosity why the no of spaces will be always one less than no of depostion points. I know if i draw a pic then it gets proved but if i want to find out the reason mentally my intution does not help me.
Just a analogy i tried to draw

A-------B -->if i want to divde into 2 parts A--|--B
<--6km-->                                   <-->C<--> "C is the point where we divide 6km into 2 spaces/parts"
                                            3km   3km

So i find that when 6 is divided by 2 it give rises to 3 points (A,B,C) so in my way I found out no of spaces will be always one less than no of depostion points(A,b,c)

twist at the end, as some people, not thinking deeply enough, would just divide 10 km by 11

I see that Common difference between each term=(total gap between frst and last term)/(no of terms -1). Does the derivation of this formula have something to do with S= 0.5 n [ 2a + (n-1) d ]

BUt seriously when i was finding common difference while i was doing the sum in my way, this formula never came to my mind . I dont know what should be my thinking process to derive or understand the formula ; other than drawing it out and then checking it out .

WHY DO U SAY "not thinking deeply enough, would just divide 10 km by 11" Why would someone divide by 10km/11..My intution suggest that division of FRACTION gives me the worth of each part , here worth MEANS common gap between every 2 points right?

11 is the no of deposits ..now i discover that if i want to divide 10 km (total distance between 2 points) by a value , there will be 2 things to define(one is denominator and the other is quotient) denominator have to be total no of gaps that i want 10km to divide into and the common worth of each gap.
Now what information i have till this ? i know distance is 10km , no of trips or deposits=11
my equation stands like this => Common worth of each gap=10km divided by (total no of gaps)

Now the issue is i dont know what is the total no of gaps that will fit in those 10km among 11 deposits. If i draw a picture, I know from the picture that I have one less no of gaps than no of deposits but without picture how will I think about "n-1"

But then you are taking n to mean only half the number of one-way trips, so your answer would be wrong! I suppose you are thinking of n as the number of deposits, not the number of trips. You never said what n meant.

I took N as no of 1 way trips.

550/2=n/2∗(20+30)⟹n=11.
n gives me the no of deposits and the no of trips.
@Dr.Peterson

 

[JeffM]

If d is the distance between each drop site, then each one way trip is d longer than the one before EXCEPT FOR THE FIRST TRIP, which is just 20.
So the number of times you need to add d is (n - 1).

 

[Dr.Peterson]

WHY DO U SAY "not thinking deeply enough, would just divide 10 km by 11" Why would someone divide by 10km/11..My intution suggest that division of FRACTION gives me the worth of each part , here worth MEANS common gap between every 2 points right?

I said this because many students don't take time to think. You did, and got it right without having to explain. That's all that needs to be said.

Your intuition, or a picture, is all you need to explain it.

I took N as no of 1 way trips.

550/2=n/2∗(20+30)⟹n=11.
n gives me the no of deposits and the no of trips.

No, as I mean it, there are two one-way trips (there, and back) for each deposit. What you presumably mean here is that you took n as the number of trips from the quarry to a deposit, and just ignored the trips back. So your n is the same as the number of two-way trips. You are just using words differently than I did.

 

[Saumyojit]

If d is the distance between each drop site, then each one way trip is d longer than the one before EXCEPT FOR THE FIRST TRIP, which is just 20

@JeffM I understood that Term subscript (n)=Term subscript (n-1)+ common difference [Next term=Previous term + cd] and the first term has to be given along with cd to get the next 'n' terms.

So the number of times you need to add d is (n - 1)

I understand if i add d (n-1) times=d*(n-1) I will get the whole gap between first and last term.
sir JEff did u gave me ur reply in post 12 to my doubt of "why the no of spaces will be always one less than no of depostion points. I know if i draw a pic then it gets proved but if i want to find out the reason mentally my intution does not help me"

 

[JeffM]

sir JEff did u gave me ur reply in post 12 to my doubt of "why the no of spaces will be always one less than no of depostion points. I know if i draw a pic then it gets proved but if i want to find out the reason mentally my intution does not help me"

Yes

 

[Saumyojit]

each one way trip is d longer than the one before EXCEPT FOR THE FIRST TRIP

It feels like I understand some part but still having a doubt.
I hope i am interpreting right --> when u are saying EXCEPT FOR THE FIRST TRIP U try to tell me the common difference in forward Arithmetic progression does not produce the first term , although if i know the first term i can get all the other n-1 terms at each step by adding previous terms to d. Which i agree .

See if there are 4 terms (A,b,c,d)--> gap no1 connects (A,b); gap no 2 connects (b,c); gap no3 connects (c,d). GAP NO 1 i.e (common d ) does not exclude the first term in terms of Connecting or relating between first and the next term.

But in ur explantion U are trying to tell gap no1(cd) fails to produce A but it is helping to produce B which carries forward to produce all the remaining terms which i understood BUT I Cannot link the relation of my "why N-1?" with ur explanation.
I KNOW Total no of cd=total no of gaps .

CONCLUSION:I am sensing the total no of gaps or total no of cd is 1 less thats because after the first term value i know I can get the value of the remaning n-1 terms by using n-1 times d .n-1 means total no of spaces which is 10 required to produce the next 10 terms .

Is this the reason ? If u can explain it much more intutively it would be helpful. @JeffM

 

[JeffM]

It feels like I understand some part but still having a doubt.
I hope i am interpreting right --> when u are saying EXCEPT FOR THE FIRST TRIP U try to tell me the common difference in forward Arithmetic progression does not produce the first term , although if i know the first term i can get all the other n-1 terms at each step by adding previous terms to d. Which i agree .

See if there are 4 terms (A,b,c,d)--> gap no1 connects (A,b); gap no 2 connects (b,c); gap no3 connects (c,d). GAP NO 1 i.e (common d ) does not exclude the first term in terms of Connecting or relating between first and the next term.

But in ur explantion U are trying to tell gap no1(cd) fails to produce A but it is helping to produce B which carries forward to produce all the remaining terms which i understood BUT I Cannot link the relation of my "why N-1?" with ur explanation.
I KNOW Total no of cd=total no of gaps .

CONCLUSION:I am sensing the total no of gaps or total no of cd is 1 less thats because after the first term value i know I can get the value of the remaning n-1 terms by using n-1 times d .n-1 means total no of spaces which is 10 required to produce the next 10 terms .

Is this the reason ? If u can explain it much more intutively it would be helpful. @JeffM

You do not need to add anything to get the first value. After that first value, you need to add d to the previous value.

So if there are 11 locations numbered 1 through 11, you don't need to add anything to get the value of location 1.

 

[Saumyojit]

You do not need to add anything to get the first value. After that first value, you need to add d to the previous value.

So if there are 11 locations numbered 1 through 11, you don't need to add anything to get the value of location 1.

I will explain it to myself "why the no of spaces will be always one less than no of depostion points " like this because total no of common diffrences or total no of spaces is required to get multipled by value of C.D to produce next n-1 or next 10 terms thats why total no of spaces will be one less than no of term.

 

[JeffM]

I will explain it to myself "why the no of spaces will be always one less than no of depostion points " like this because total no of common diffrences or total no of spaces is required to get multipled by value of C.D to produce next n-1 or next 10 terms thats why total no of spaces will be one less than no of term.

Any way that works for you is a good way

 

[Saumyojit]

@Dr.Peterson THAT MEANS if i take one way trip : every trip defination is--> starting from the quaary where it picks up load and ends where it deposits
EVERY TRIP STARTS FROM QUARRY and ends at deposition point if it is one way or ends at quarry if it is ROUND trip.

mostly because that is the same as the number of deposits

One trip gives me one depostion point . Distance covered in each one way trip gives me the distance of quarry to that point.

at FIRST i did not realize that DISTANCE from quarry to deposit is actually each term of this AP in forward progression

When i first saw the question "find the no of trips" I knew that each term means each deposit points but did not REALIZE that finding no of each trip will give total no of deposition points and finding the value of each 1 way Trip will give me the value of each deposit which in turn gives me the value of Every Term of Ap in forward progression .

AT FIRST I could not RELATE "trips" means "TERMS" in this sum.