A wire of length 6 m is cut successively into 25 portions such that each portion is 1 cm longer than the preceding portion. If the whole wire is used up in the cutting, find the length of the first portion cut from the wire.

Question

debbieg · Answer

Welllll.... you can set up the sum:

$$\Large 6=\sum_{n=0}^{24}(x+n)$$

Now, if you think about that sum, how can that be simplified? 

$$\Large \sum_{n=0}^{24}(x+n)=x+0+x+1+x+2+...+x+24$$
$$\Large =(??)x+\sum_{n=0}^{24}n$$

So tell me what goes into the (??).... ?

And, if I'm not mistaken, there is a general formula for $\Large \sum_{n=0}^{24}n$ right?

So you can get an equation in x that you can solve. :)

debbieg · Answer

hold on... make that 600 on the LHS, since the "1" on the RHS is in cm, need the LHS also in cm... my bad. :)

$\Large 600=\sum_{n=0}^{24}(x+n)$

anonymous · Answer

If the initial term of an arithmetic progression is a_1 and the common difference of successive members is d, then the nth term of the sequence (a_n) is given by:
a-n=a1+(n-1)d --> a-25=a-1+(25-1)*1 -->a-25 - a-1=24 (A) 
The sum of the members of a finite arithmetic progression is 
S-n=n/2(a_1+a-n) --> 600(cm)=25/2(a-1+a-25) --> a-1+a-25=48 (B)

Now using both A and B we can set a 2-unknown set of parameters a-1 and a-25 as

a-25 - a-1=24
a-25 + a-1=48
solving for a-1, we get
2a-1=24 --> a-1=12