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    (Original post by HeatherChandler)
    T_n = a + (n - 1)d

    S_n = \frac{n}{2}(2a + (n-1)d)

    T_4 and S_6 give you simultaneous equations with which you can find a.
    *in a small voice* But don't I have 3 unknowns?
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    No you have two - a & d. n is not unknown because it says the sum to six terms...
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    What about k?!!
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    You need to find a & d first - Ignore this

    EDIT: Sorry you just work it out normally. The question doesn't say find k
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    So how do I solve

    7k + 9 = 6a + 15d and
    3k = a + 3d



    ?

    Sorry for being so.. rubbish.

    OH! I see..

    Wow. I'm worse at maths than I ever thought possible! I'll stop pestering you now.. thanks loads though
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    Use either elimination or substitution

    Spoiler:
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    Substitue a=3k-3d into eqn 1
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    (Original post by Avalar)
    Wow. I'm worse at maths than I ever thought possible! I'll stop pestering you now.. thanks loads though
    You should absolutely pester if you don't understand something.
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    Helped me out too. Thanks. I was on the same question lol, good timing
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    Sum= n/2 (2a+(n-1)d)
    675 = 10/2( 2a + (10-1)d)
    675= 5(2a+9d)
    2a+9d=675/5
    2a+9d=135

    this is equation one

    since the series is arranged from ascending order,
    the first term(shortest side) is (a) and as in the question the longest side or the last time is twice that of the shortest side
    so, shortest side is "a"
    and longest side is "2a"

    so the other equation is
    last term = 2a
    in other words,
    a+(n-1)d = 2a
    a+(n-1)d - 2a =0
    since n is 10
    a+(10-1)d -2a=0
    a+9d-2a=0
    -a+9d=0

    so now we can solve the simultaneous equation


    2a+9d=135 (equation obtained before)
    -a+9d=0

    the ans is a=45 and d=5
    so the first term is 45 and common difference is 5.
    hope u got it
 
 
 
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