1. ## 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!

2. 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: $u_n = a+(n-1)d$

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

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

iii) Solve for $n$ using the sum formula given $u_1$ and $d$

3. 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)

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

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

(ii)

As $S_n=5000$

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

from which the first term can be deduced.

(iii)

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

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