Arithmteic series

Jaffa · #1 $Old$ October 4th, 2008, 06:55 AM

Can somebody help me with this question,
From a piece of wire 5000mm long, n pieces are cut,each piece 10mm longer
than the previous piece. If u is the lenght of the i-th piece and Si is the
cumulative lenght of i pieces.

(i) Express Un and Sn in terms of Ui and n
(ii)Determine the number of pieces that can be cut if Ui is 155mm

Could you please explain the answer

thanks

bkarpuz · #2 $Old$ October 4th, 2008, 09:13 AM

Quote:

Originally Posted by Jaffa $View Post$

Can somebody help me with this question,
From a piece of wire 5000mm long, n pieces are cut,each piece 10mm longer
than the previous piece. If u is the lenght of the i-th piece and Si is the
cumulative lenght of i pieces.

(i) Express Un and Sn in terms of Ui and n
(ii)Determine the number of pieces that can be cut if Ui is 155mm

Could you please explain the answer

thanks

I will show you some little about the solution.
(i) Let $u(1)$ be the length of the first term.
Then, we have $u(2)=u(1)+10$ , $u(3)=u(2)+10=u(1)+20$ , and in general $u(i)=u(i-1)+10=u(i-2)+20=\ldots=u(1)+(i-1)10$ for $i=2,3,\ldots,n$ .
And
$S(i)=\sum\limits_{j=1}^{i}u(j)$
....... $=\sum\limits_{j=1}^{i}[u(1)+(j-1)10]$
....... $=u(1)\sum\limits_{j=1}^{i}1+10\sum\limits_{j=1}^{i}(j-1)$
....... $=iu(1)+10\frac{(i-1)i}{2}$
....... $=iu(1)+5(i-1)i$
for $i=1,2,\ldots,n$ .
(ii) I guess $u(1)=155mm$ , just consider the arguments in (i), and solve $n$ by setting $S(n)=5000mm$ .