Maths help

[Tony Starks]

I've just started a correspondence maths course and I'm stuck on how this example works. You're all cleverer than me so how does this work?



Its to do with Arithmetic Sequences and I can work out what the number for a sequence will be using the formula tn = a+(n-1)d to work out say the 9th term would be this value. I'm just stumped on how to work it out from being given the value to work out the sequence number.

[otherman]

It's all right there. You just subsitute letters for numbers you know, then simplify til you get the answer.

You're given the equation and told that tn=138 and that a=90 and d=4. so first, put those into the equation.

Then open up (N-1)x4 to 4N-4. Put in any value of N at all and you'll see that this works

Now do 90-4 = 86

Gradually we're reducing the number of digits we have to deal with.

Now subtract 86 from both sides (so we stay equal both sides of the = sign)

All that's left is 52=4N

So divide both sides by 4 and 13=N.

[Tony Starks]

So, they should have left a space between the first 2 lines and the last 4?

Now, you'll have to excuse my terminology and understanding here. But the first row (138 = 90+4n-4), I'm largely ignoring 4n in the first instance as its just doing 90-4 = 86 which I'm then subtracting from my original number of 138 to give 52.
The 4n is then the term number divided by 4, which is then our sequence number of 13

[Tony Starks]

otherman said:

0
Then open up (N-1)x4 to 4N-4. Put in any value of N at all and you'll see that this works

.

This is whats confusing me

[otherman]

Tony Starks said:

otherman said:

0
Then open up (N-1)x4 to 4N-4. Put in any value of N at all and you'll see that this works

.

This is whats confusing me

(N-1) x 4

You can expand expressions like this by multiplying the N by 4, and the 1 by 4. So it's 4N-4

Say N=3
(N-1) = 2.
2 * 4 = 8

So the value of (N-1)*4 is 8.

Now let's try
4N-4
4*3 - 4 = 8

Same answer. Use any value for N and you'll find that (N-1)x4 is the same as 4N-4.

[Tony Starks]

Nice one thanks

[StuntmanMike]

Christ! Are you lot speaking Klingon?

[Tony Starks]

One more quick one,

On the 86 + 4n part, is there a rule to say whether its 4 multiplied/divided by n?

StuntmanMike said:

Christ! Are you lot speaking Klingon?

Pretty much, I left school nearly 30 years with a D grade in maths, so it's pretty hard. I'm foolishly doing a physics course too lol.

One thing we all say is 'I'll never need algebra', but it's all around you. My whole work uses CNCs the run off it. I use the pythagoras therum to work out how measurements to see if things are square.

[TwigtheWonderkid]

Tony Starks said:

One thing we all say is 'I'll never need algebra', but it's all around you.

Indeed. Every time I look at my x, I wonder y.

I remember saying to my maths teacher at school that I'll never need to use algebra once I leave. And she said "you're absolutely right....but the smart kids might".

[Tony Starks]

TwigtheWonderkid said:

Indeed. Every time I look at my x, I wonder y.

I remember saying to my maths teacher at school that I'll never need to use algebra once I leave. And she said "you're absolutely right....but the smart kids might".

Ouch

[weeredmetro]

Tony Starks said:

One more quick one,

On the 86 + 4n part, is there a rule to say whether its 4 multiplied/divided by n?

4n means 4 times n. 4/n would be 4 divided by n

The multiplication sign is probably the most "taken for granted" once you are past school level maths.

At school, you are taught +, -, × and ÷

+ and - are fairly universal, but 3/2 for 3÷2 is very common and for multiplication, the × might be written as the letter X, or as a • which frequently mutates into a . at which point it can be confused for a decimal point. Thanks to Microsoft, in Excel you have to use * for multiplication which has also made its way into mainstream use.

5(n+1) is therefore the same as
5×(n+1)
5•(n+1)
Colloquially 5*(n+1)
Typographically annoyingly 5.(n+1) and 5x(n+1)

Isn't standardisation of symbols fun?

[Tony Starks]

I'm so glad I dont need to program the machines I use

Right, this has me stumped. I cant seem to get the the answer to work out correctly.

Josh has started jogging to keep fit. On the first day he jogs 3 kilometres. Each day he adds 0.5 kilometres to the length of his jog. Josh plans to keep increasing the distance he jogs until he gets to jog 8 kilometres.
How many days will it take for Josh to complete one jog of 8 kilometres?

So I've been using the formula for an Arithmetic Series :
Sn = n/2 [2a + (n-1)d]

As well as the tn = a+(n-1)d formula.

If I write it down, the bow legged muppet will take 11 days to reach 8km.

If I use the tn formula it comes out as 6.5 days to reach 8km.

Edited by Tony Starks on Sunday 7th February 04:11

Edited by Tony Starks on Sunday 7th February 20:47

[OzzyR1]

11 days would seem to be the obvious answer but must qualify my opinion with the fact I am useless at these sort of questions.

[Gary C]

so where N = number of days

3+(0.5 * (N-1)) = 8 (3 days to start plus number of days multiplied by half a km but day 1 adds nothing so N-1)

3+(0.5*N - 0.5) = 8 (open up inner bracket)

0.5*N - 0.5 = 5 (take 3 from both sides)

0.5 * N = 5.5 (add 0.5 to both sides)

N = 11 (divide both sides by 0.5)

Think this works (im crap at arithmetic despite having a degree in engineering )

Edited by Gary C on Sunday 7th February 04:28

[Tony Starks]

Gary C said:

so where N = number of days

3+[b](0.5 *[b/] (N-1)) = 8 (3 days to start plus number of days multiplied by half a km but day 1 adds nothing so N-1)

3+(0.5*N - 0.5) = 8 (open up inner bracket)

0.5*N - 0.5 = 5 (take 3 from both sides)

0.5 * N = 5.5 (add 0.5 to both sides)

N = 11 (divide both sides by 0.5)

Think this works (im crap at arithmetic despite having a degree in engineering )

Edited by Gary C on Sunday 7th February 04:28

you seem to have multiplied here, when the equation is Add? unless theres something I'm missing

[Ultuous]

Using the 'tn' formula:

tn=a+(n-1)d

Insert the known values:

8 = 3 + (n-1)*0.5

Rearrange to make n the subject of the formula as follows...

Take 3 from both sides:

5 = (n-1)*0.5

Divide both sides by 0.5:

5/ 0.5 = (n-1)

Add 1 to both sides:

5/0.5 + 1 = n --> n = 5/0.5 + 1= 5 *2 + 1 = 10 + 1 = 11

[Tony Starks]

Got it, I was assuming 8 was 'n', when in actual fact I technically didn't know this figure. Once I did it like Otherman explained it all fell into place. 'n' was actually 11 - the number of days.

I just need a better understanding of the questions and explanations.

[BelfastBlack]

Ultuous said:

Using the 'tn' formula:

tn=a+(n-1)d

Insert the known values:

8 = 3 + (n-1)*0.5

Rearrange to make n the subject of the formula as follows...

Take 3 from both sides:

5 = (n-1)*0.5

Divide both sides by 0.5:

5/ 0.5 = (n-1)

Add 1 to both sides:

5/0.5 + 1 = n --> n = 5/0.5 + 1= 5 *2 + 1 = 10 + 1 = 11

You've got to the right answer but it's arse about face. You've got to remember your order if operations (a term I haven't heard in about 15 years but just came flooding back).

OP, The easiest way to learn maths is to do loads of examples using the same method but different numbers. Then you lock in the method and the numbers/letters don't matter.

[Ultuous]

BelfastBlack said:

You've got to the right answer but it's arse about face. You've got to remember your order if operations (a term I haven't heard in about 15 years but just came flooding back).

Do you care to elaborate? Genuinely keen to understand what specifically you'd do differently!

[Gary C]

Tony Starks said:

you seem to have multiplied here, when the equation is Add? unless theres something I'm missing

I did multiply 0.5 by N because there would be multiple days (N) of 0.5km to add to the starting number of 3km

so it was 3 km to start plus 0.5km multiplied by an unknown amount of days called N that results in 8 days

so 3km + (0.5km x N) = 8 days

Except that the first day didn't add any extra so take 1 away from N

so 3 + (0.5 x (N-1) ) = 8