# Arithmetic Series Help

• Mar 29th 2010, 02:09 PM
JP103
Arithmetic Series Help
First of all I'm not sure if this is the right section, so I apologise if it's not.

I was hoping that someone could help me out with the following question:

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

Express Un and Sn in terms of U1 and n.
• Mar 29th 2010, 02:35 PM
Anonymous1
$U_n = u_1 + 10(n-1)$

$S_n = \sum_{i=1}^n U_i = \sum_{i=1}^n [u_1 + 10(i-1)]$ $= u_1 + 10\sum_{i=1}^n i -\sum_{i=1}^n 1 = u_1 + 10\frac{n(n+1)}{2} - n = u_1 +n(5n + 4)$

Also, $S_n = 5000 \Rightarrow u_1 = 5000 - 5n^2 - 4n.$
• Mar 30th 2010, 03:40 AM
JP103

However the others don't seem to add up, I had to find U1 as the second part of the question and I've calculated it as being 80 which I've checked and it is right. If you put 80 into your formulae it doesn't equal 5000.

Thank you anyway though bud.
• Mar 30th 2010, 06:16 AM
Soroban
Hello, JP103!

Anonymous1 made some small errors. . .

$S_n \;=\; \sum_{i=1}^n U_i$

. . $=\; \sum_{i=1}^n \bigg[U_1 + 10(i-1)\bigg]$

. . $=\; \sum_{i=n}^n\bigg[U_1 + 10i - {\color{red}10}\bigg]$

. . $= \;\sum_{n=1}^nU_1 + 10\sum_{n=1}^ni - 10\sum_{n-1}^n1$

. . $=\;{\color{red}n}U_1 + 10\frac{n(n+1)}{2} - 10n$

$\boxed{S_n \;=\;nU_1 + 5n(n-1)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Using the sum-formula: . $S_n \:=\:\frac{n}{2}\bigg[2a + (n-1)d\bigg]$

. . we have: . $S_n \;=\;\frac{n}{2}\bigg[U_1 + (n-1)10\bigg] \quad\Rightarrow\quad S_n \;=\;nU_1 + 5n(n-1)$