hmm what do you think?

ads114 · #1 $Old$ November 5th, 2009, 05:18 AM

hi i am going the right way about this question?

from a wire 5000mm long,n pieces are cut,each piece is 10mm longer than the previous one.
if ui is the i^th piece and Si is the total length of i pieces.

1.express un and Sn in terms of u1 and n.

2.having cut 25 pieces from the wire find the length of u1.

so i am thinking.....

u1 is the length of the 1st term

u2=u1+10mm

u3 = u2+10mm and so on but thats where i am getting mixed trying to get this into some sort of further arithmetic series

it may be something like ui-1+10.................

stuck.

i am far off?

CaptainBlack · #2 $Old$ November 5th, 2009, 06:27 AM

Quote:

Originally Posted by ads114 $View Post$

hi i am going the right way about this question?

from a wire 5000mm long,n pieces are cut,each piece is 10mm longer than the previous one.
if ui is the i^th piece and Si is the total length of i pieces.

1.express un and Sn in terms of u1 and n.

2.having cut 25 pieces from the wire find the length of u1.

so i am thinking.....

u1 is the length of the 1st term

u2=u1+10mm

u3 = u2+10mm and so on but thats where i am getting mixed trying to get this into some sort of further arithmetic series

it may be something like ui-1+10.................

stuck.

i am far off?

There seems to be something missing from this question, but:

$u_k=u_1+10(k-1)$

I also have my doubts about if number theory is the appropriate forum for this, to me it looks like pre-calculus.

CB

ads114 · #3 $Old$ November 16th, 2009, 02:36 AM

Quote:

Originally Posted by CaptainBlack $View Post$

There seems to be something missing from this question, but:

$u_k=u_1+10(k-1)$

I also have my doubts about if number theory is the appropriate forum for this, to me it looks like pre-calculus.

CB

hi there ok i thnk i have managed to get an expression for Sn and Ui.

i think its Ui =U1 +10(i-1).

Sn =i/2(2U1+(i-1)10)?

assumimg that is correct i now need to transform for U1.

i had a go at this and got eventually this

2Sn=i(2U1+(i-1)10).
multiplied the brackets 1st ending up with the following

U1=2Sn-10i/2i.? does this transformation look ok?

CaptainBlack · #4 $Old$ November 16th, 2009, 02:46 AM

Quote:

Originally Posted by ads114 $View Post$

hi there ok i thnk i have managed to get an expression for Sn and Ui.

i think its Ui =U1 +10(i-1).

Sn =i/2(2U1+(i-1)10)?

$S_n$ cannot depend on $i$ :

$S_n=\sum_{i=1}^n U_i=nU_1-10n+10\sum_{i=1}^n i$

CB

HallsofIvy · #5 $Old$ November 16th, 2009, 06:26 AM

Any arithmetic progression has the nice property that the "average" value of [math]a_n, a_{n+1}, \cdot\cdot\cdot, a_m[/itex] is the average of the two end vaues, $(a_n+ a_m)/2$ . Here, the common difference is 10 so $u_1= u_1+ 10(i-1)$ . The last, "n"th piece, has length $u_n= u_1+ 10(n-1)$ . The average of the first and last piece, and so the average length of all pieces, is [math](u_1+ u_1+ 10(n-1))/2= u_1+ 5(n-1)[/itex]. The length of all the pieces together, then, is $n(u_1)+ 5(n-1))$ and that must be equal to the length of the original piece of wire, 5000: $u_1n- 5n^2- 5n= 5000$ . You can solve that for $u_1$ as a function of n and so get $u_i$ as a function of i and n. You can use the same "averaging" idea to get the total length of the first i pieces as a function of i and n.