Recent content by mattyh3

  1. M

    Solving for n in a Summation Series

    so 2Sn= n2a + dn-d am i on the right lines or still off
  2. M

    Solving for n in a Summation Series

    or sn= n+ 2a +dn-d/2 then sn= 2a+d2n-d/2 think am going totally wrong as am not to sure what the next step is etc
  3. M

    Solving for n in a Summation Series

    but am guessing with 2dn-2dn they would cancel each other out
  4. M

    Solving for n in a Summation Series

    still confused as to what's happeneds to the Sn so to me the next step would be 2a + 2dn-2dn or am i wrong
  5. M

    Solving for n in a Summation Series

    its knowing/ remembering how to do that as well. am really struggling here :( as mine seems a lot harder to do than what is in step 6
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    Solving for n in a Summation Series

    i have found on google how to do what is in 6 but looking at that i aint got a clue on where to start with getting n as the subject in this sn=n/2 (2a+(n-1)d).. any help would be great just to get me started even a first step or so as its only an example in my book and i just want to be able...
  7. M

    Solving for n in a Summation Series

    i have studied them but many years ago and i have seen that before but that's why am struggle and i can't seem to transpose that equation to find n and i have tried loads of times and i end up with answers way off
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    Solving for n in a Summation Series

    i haven't solved it the lesson book did the equation and at the end said n=12.. but what am trying to say is i don't know how its transposed the equation to make n the subject and to come up with n=12 i was wondering if someone could fill in the missing steps as to how it arrives with n=12
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    Solving for n in a Summation Series

    yeah i know the answer is 12 because it tells me... but i am not sure how it is coming up with 12 as i am unsure on how to make n the subject so i can work it out myself as i need to understand how for assignment question later on
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    Solving for n in a Summation Series

    Homework Statement How many terms of the AP, 12, 16, 20, 24 . . . . . must be taken for the sum to equal 408? Solution For this series, a = 12, d = 4 and Sn = 408. its asking me to solve for n and gives me the answer to n as 12 at the end but when i try to calculate it myself i don't get 12...
  11. M

    Engineering How Do You Calculate Apparent and Reactive Power in AC Circuits?

    thanks for the help just need to find how to do complex calcs on my calculator now
  12. M

    Engineering How Do You Calculate Apparent and Reactive Power in AC Circuits?

    o great i thought i was wrong lol... so now i just need to take off the 10 ohm resistance from the real value of 19.40828158597505 to get z= 9.408 ohms
  13. M

    Engineering How Do You Calculate Apparent and Reactive Power in AC Circuits?

    thanks for the help but i can't even use the calculator it gives me weird answers nothing like what's on here this is what i get 19.408277013161786 + i*19.70643553618951
  14. M

    Engineering How Do You Calculate Apparent and Reactive Power in AC Circuits?

    if i divide them into 120 it get 38.402+39.843j i don't understand any other way of doing it as am not great with maths
  15. M

    Engineering How Do You Calculate Apparent and Reactive Power in AC Circuits?

    still can't find anything of any use that i can understand where the 120 goes and you end up with 19.9079+19.1882j
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