anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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DebbieG
  • DebbieG
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
  • DebbieG
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
  • anonymous
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

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