I don't know what to do with this question, please help!
A piece of string 10m long is cut into pieces, so that the lengths of the pieces form an arithmetic sequence.
(i) The lengths of the longest and shortes pieces are 1m and 25cm respectively; how many pieces are there?
Answer:16
(ii) If the same string had been cut into 20 pieces with lengths that formed an arithmetic sequence, and if the length of the second longest had been 92.5 cm, how long would the shortest piece have been?
(ii) Is the one I don't know what to do with. I have already completed (i)

- anonymous

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

which one !?

- anonymous

I need help with the last question please.

- welshfella

Arithmetic Series:
2nd term = a1 + d where a = first term and d = common difference
so here
92.5 = a1 + d

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- welshfella

the sum of 10 terms = the length of the string = 10,000 cms

- anonymous

Um isn't 1000cm?

- welshfella

Now the sum of n terms is
(n/2)[2a1 + (n - 1)d]
so here n = 20 and Sum = 10,000 so we have
10,000 = 10[2a1 + 19d]

- welshfella

so we have the system of equations
a1 + d = 92.5
20a1 + 19d = 10,000

- anonymous

Ok, one little question- why is it 2a1 ?

- welshfella

2 * first term
a1 will be the first term = shortest piece

- anonymous

Oh ok

- welshfella

so you need to find the value of a1

- anonymous

I use a different form of that equation normally so was confused for a second

- welshfella

if you multiply the first equation by -19 and add you can find a1

- welshfella

yes a1 is usually used in the US
in UK we usually use just a

- anonymous

a1 + d = 92.5 sorry, could you just explain more about how you got this part? I dont really understand why the first term plus the common term equals the second last term? I know I am missing something

- welshfella

oh - maybe i was wrong there sorry - -i took that to be the second term
- its the 19th term

- welshfella

also the total length = 1000 (wrong again!)

- anonymous

Haha, it happens! No worries.

- welshfella

so the 2 equations are
a1 + 18d = 92.5
20a1 + 19d = 1000

- anonymous

Ok, I am not sure I understand the concept behind that second equation

- welshfella

that is the formula for the sum of n terms of an arithmetic series
Sn = (n/2)[2a1 + (n - 1)d]
here Sn = length of the string = 1000 cms, n = 20 (20 pieces) , a1 = first term, d = common difference

- anonymous

Ok. Then I am meant to simultaneously solve those two equations?

- welshfella

yes

- welshfella

i bit messy Have you got a graphical calculator?

- anonymous

No I don't. Oh well, I will try it quickly now without one

- amistre64

is our shortest piece still 25cm?

- anonymous

No I don't think so. That was for the first question

- amistre64

how long should shortest piece ... forgot to read thru it all the way

- welshfella

hmm the results don't seem to give rational values..

- anonymous

Yeah I am not getting anything that makes sense.

- anonymous

I got d=2.5 a=47.625 after rounding

- welshfella

i thnk the equations are correct..

- anonymous

I probably just did something wrong then.

- welshfella

yes the calc gives 47.631 and 2.4927

- amistre64

10m long
20(2a+19d)/2 = 1000
a+18d = 92.5
a is about 2.5, and d is about 5

- welshfella

yea
my second equation was wrong
10(2a + 19d) = 201 + 190d

- amistre64

2a+19d = 100
a+18d = 92.5
2a +19d = 100
-19/18a -19d = -19/18(92.5)
a = (100-(19/18(92+1/2)))/(2-(19/18))

- welshfella

2.5 for shortest length
and 5 for common difference

- welshfella

it was 190d not 19d

- anonymous

Ok..

- welshfella

do you see the mistake
I didnt distribute 10 over the (2a + 19) correctly

- welshfella

I thought someone would spot my deliberate mistake
- oh amistre did!

- anonymous

Oh! I see now!

- welshfella

good
there is another formula for sum of n terms of an arithmetic series
Sn = (n/2)[ a1 + L] where L is the last term
If you know the last term its a lot simpler

- anonymous

Yes, thats the form I normally use and I didn't remeber about the other one not needing the last term which is a big part of why I got stuck I think
Just trying to solve those two equations myself now.

- welshfella

probably best to multiply first one by -20 and add to get d first

- anonymous

Yup! Did it and got the two right answers! Thank you so much for your help!

- welshfella

yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.