# Arithmetic series.

• Dec 20th 2010, 09:09 AM
DanBrown100
Arithmetic series.
Not sure if this is the right section to be posting in but here goes.

From a piece of wire 5000mm long, n pieces are cut, each being 10mm longer than the previous piece. If ui is the length of the ith piece and Si is the cumulative length of i pieces:

(i) Calculate un and Sn in terms of u1 and n
(ii) If exactly 25 pieces can be cut from the wire, determine the length of u1
(iii) Determine the number of pieces that can be cut if u1 is 155mm

Any pointers or help would be very welcome guys!
• Dec 20th 2010, 09:50 AM
e^(i*pi)
This is an arithmetic sequence problem, not a factor theorem one

Note that there is a common difference between each length of wire

i) In other words this is an arithmetic sequence with first time u1 and common difference 10: $\displaystyle u_n = a+(n-1)d$

Similarly $\displaystyle S_n = \dfrac{n}{2}(2a+(n-1)d)$

ii) find $\displaystyle u(25)$ and solve for $\displaystyle u_1$

iii) Solve for $\displaystyle n$ using the sum formula given $\displaystyle u_1$ and $\displaystyle d$
• Dec 20th 2010, 04:03 PM
Quote:

Originally Posted by DanBrown100
Not sure if this is the right section to be posting in but here goes.

From a piece of wire 5000mm long, n pieces are cut, each being 10mm longer than the previous piece. If ui is the length of the ith piece and Si is the cumulative length of i pieces:

(i) Calculate un and Sn in terms of u1 and n
(ii) If exactly 25 pieces can be cut from the wire, determine the length of u1
(iii) Determine the number of pieces that can be cut if u1 is 155mm

Any pointers or help would be very welcome guys!

Yes,

this is an arithmetic sequence and series.

(i)

$\displaystyle u_n=u_1+(n-1)10$

$\displaystyle \displaystyle\ S_n=\frac{n}{2}\left[2u_1+(n-1)10\right]$

(ii)

As $\displaystyle S_n=5000$

$\displaystyle 5000=\displaystyle\frac{25}{2}\left[2u_1+240\right]$

from which the first term can be deduced.

(iii)

$\displaystyle u_1=155\Rightarrow\frac{n}{2}\left[310+(n-1)10\right]=5000$

$\displaystyle 10000=300n+10n^2\Rightarrow\ n^2+30n-1000=0\Rightarrow\ (n+50)(n-20)=0$