# arithmetic series

• Dec 8th 2010, 03:11 PM
sigma1
arithmetic series
A piece of cord is of length 64 m. It is cut into 16pieces whose lengths are in arithmetic
progression. The length of the longest piece is three times that of the smallest. Find the length of the shortest piece of chord.

$\displaystyle S_n = 64$

length of shortest piece is $\displaystyle x$
length of longest piece is $\displaystyle 3x$

is it right to assume that the shortest piece is the first term and the largest piece is the last term.

is this correct?
i have formed two equations

$\displaystyle 3x = x+(15)d$ equating the longest piece to the last term

$\displaystyle 64 = 16/8 (x + x+15d)$ equating the total length of cord.
• Dec 8th 2010, 03:47 PM
DrSteve
Why are you dividing by 8? I believe you should be dividing by 2. Also, replace x+15d by 3x in the second equation.
• Dec 8th 2010, 03:51 PM
pickslides
Here's some hints to get you started

$\displaystyle S_{n} = n\times \frac{a+t_n}{2}$

$\displaystyle S_{16} = 64$

$\displaystyle a+(a+d)+(a+2d)+\dots +(a+15d)=64$
• Dec 8th 2010, 03:56 PM
Soroban
Hello, sigma1!

Your set-up is correct . . . except for that "8".

Quote:

$\displaystyle \text{A piece of cord is of length 64 m.}$
$\displaystyle \text{It is cut into 16 pieces whose lengths are in arithmetic progression.}$
$\displaystyle \text{The length of the longest piece is three times that of the smallest.}$
$\displaystyle \text{Find the length of the shortest piece of cord.}$

$\displaystyle S_n = 64$

$\displaystyle \text{length of shortest piece is }x$
$\displaystyle \text{length of longest piece is }3x$

$\displaystyle \text{Is it right to assume that the shortest piece is the first term}$ . . $\displaystyle \text{and the largest piece is the last term?}$

$\displaystyle \text{i have formed two equations:}$

$\displaystyle 3x \:=\: x+(15)d\qquad \text{equating the longest piece to the last term}$

$\displaystyle 64 \:= \frac{16}{2}(2x +15d) \qquad\text{equating the total length of cord}$

$\displaystyle \begin{array}{cccccccc} \text{Your 1st equation is:} & 2x - 15d &=& 0 & [1] \\ \text{Your 2nd equaton is:} & 2x + 15d &=& 8 & [2] \end{array}$

Subtract [2] - [1]: .$\displaystyle 30d \:=\:8 \quad\Rightarrow\quad d \:=\:\frac{4}{15}$

Substitute into [1]: .$\displaystyle 2x - 15(\frac{4}{15}) \:=\:0 \quad\Rightarrow\quad x \:=\:2$

Therefore, the shortest piece is $\displaystyle 2\,m.$

• Dec 8th 2010, 04:07 PM
sigma1
Quote:

Originally Posted by Soroban
Hello, sigma1!

Your set-up is correct . . . except for that "8".

$\displaystyle \begin{array}{cccccccc} \text{Your 1st equation is:} & 2x - 15d &=& 0 & [1] \\ \text{Your 2nd equaton is:} & 2x + 15d &=& 8 & [2] \end{array}$

Subtract [2] - [1]: .$\displaystyle 30d \:=\:8 \quad\Rightarrow\quad d \:=\:\frac{4}{15}$

Substitute into [1]: .$\displaystyle 2x - 15(\frac{4}{15}) \:=\:0 \quad\Rightarrow\quad x \:=\:2$

Therefore, the shortest piece is $\displaystyle 2\,m.$

ok thanks for the corrections. the 8 was a mistake i was wrighting the result instead of the 2 that should have been dividing it..