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)
which one !?
I need help with the last question please.
Arithmetic Series: 2nd term = a1 + d where a = first term and d = common difference so here 92.5 = a1 + d
the sum of 10 terms = the length of the string = 10,000 cms
Um isn't 1000cm?
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]
so we have the system of equations a1 + d = 92.5 20a1 + 19d = 10,000
Ok, one little question- why is it 2a1 ?
2 * first term a1 will be the first term = shortest piece
Oh ok
so you need to find the value of a1
I use a different form of that equation normally so was confused for a second
if you multiply the first equation by -19 and add you can find a1
yes a1 is usually used in the US in UK we usually use just a
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
oh - maybe i was wrong there sorry - -i took that to be the second term - its the 19th term
also the total length = 1000 (wrong again!)
Haha, it happens! No worries.
so the 2 equations are a1 + 18d = 92.5 20a1 + 19d = 1000
Ok, I am not sure I understand the concept behind that second equation
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
Ok. Then I am meant to simultaneously solve those two equations?
yes
i bit messy Have you got a graphical calculator?
No I don't. Oh well, I will try it quickly now without one
is our shortest piece still 25cm?
No I don't think so. That was for the first question
how long should shortest piece ... forgot to read thru it all the way
hmm the results don't seem to give rational values..
Yeah I am not getting anything that makes sense.
I got d=2.5 a=47.625 after rounding
i thnk the equations are correct..
I probably just did something wrong then.
yes the calc gives 47.631 and 2.4927
10m long 20(2a+19d)/2 = 1000 a+18d = 92.5 a is about 2.5, and d is about 5
yea my second equation was wrong 10(2a + 19d) = 201 + 190d
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))
2.5 for shortest length and 5 for common difference
it was 190d not 19d
Ok..
do you see the mistake I didnt distribute 10 over the (2a + 19) correctly
I thought someone would spot my deliberate mistake - oh amistre did!
Oh! I see now!
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
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.
probably best to multiply first one by -20 and add to get d first
Yup! Did it and got the two right answers! Thank you so much for your help!
yw
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