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Maths help please: Arithmetic Progression question

Discussion in 'General Discussion' started by Mad old tory, 10 May 2006.

  1. Mad old tory

    Sgarrista

    Joined: 25 May 2003

    Posts: 9,361

    Location: Limehouse

    Hi guys, got a question here that i can't seem to make sense

    The sum of the first 12 terms of an AP is 222, the sum of the first 5 terms is 40. Write out the first four terms of the series.

    Right I'm using te Sn=n/2[2a+(n-1)d] equation to form a simultaneous equation to work out a and d:

    222=6[2a+11d]
    40=5/2[2a+4d]

    Multiplying the brackets leaves me with:

    222=12a+66d
    40=5a+10d

    Multiplying the first one by 5 and the second by 12 to get a common a of 60 lets me find d, but it doesn't give me the right answer. for the record, a is 2 and d is 3, but i can't get those answers. I know I've obviously done something silly, or might even be doing the completely wrong thing, can someone point me in the right direction please?
     
  2. Dave

    Mobster

    Joined: 30 Oct 2004

    Posts: 4,937

    Location: Sacramento CA

    222=12a+66d
    40=5a+10d

    Is correct - using row-reduced echelon form on my TI83 gives a=2, d=3

    Long way:

    1110 = 60a + 330d (top one times 5)
    480 = 60a + 120d (bottom one times 12)

    Subtract bottom from top:
    630 = 210d

    Thus d = 3

    Substitution:
    40 = 5a +10(3)

    Thus a = 2

    Think you made some basic mistake you'll kick yourself for :)
     
  3. Mad old tory

    Sgarrista

    Joined: 25 May 2003

    Posts: 9,361

    Location: Limehouse

    yeh i have done, had 330-120 down as 110 instead of 210, can you tell I've been revising all day! :o:rolleyes::p

    Cheers for that, I knew my method was right!