Math Arithmetic Series?

anonymousanonymous
in Science & Mathematics

The question sounds like this:

"The sum of the first twenty terms of an arithmetic series is 540 and the sum of the first thirty terms of the series is 1260.

Find the first term and the common difference of the series"

I was wondering how you guys would work this out? I have tried for a long time now, and I cant quite get my head around it. I have reduced them to:

S20 = 540

540 = 20u1 + 190d

---------------

S30 = 1260

1260 = 30u1 + 435d

I have attempted to subtract the value of S20 with S30, but I got stuck form there! :/

Could you guys please help me out a little?

Thank you very much on-forehand!

Comments

  • sahsjingsahsjing

    540/20 = 27, the average of the first 20 terms

    1260/30 = 42, the average of the first 30 terms

    d = (42-27)/5 = 3, the common difference

    [a(1)+a(1)+(20-1)*3]*20/2 = 540

    a(1) = -3/2

    --------

    Get it done in the simplest way is what you always should try to learn.

  • eissaeissa

    Sn=n/2*{2a+(n-1)d} where S is the sum , n is the number of terms , a is the first term, d is the difference

    S20= 20/2*{2a+(20-1)d}=540 so ==> 540=10(2a+19d) divide by 10

    so 2a+19d=54 ...........(1)

    S30=30/2 *{2a+(30-1)d}=1260 so 15(2a+29d)=1260 divide by 15

    so 2a+29d=84 .......(2) subtract (1) from (2)

    so 10d=30 so d=30/10=3

    from (1) 2a+3(19)=54 , 2a+57=54 so a=54-57=-3

    so the first term=--3 and the common difference =3

    thanks (fawzy-maths teacher-egypt)tel 0020101238946

    [email protected]

  • mohanrao-dmohanrao-d

    20u1 + 190d = 540

    divide by 10

    2u1 + 19d = 54-----------------------eqn(1)

    1260 = 30u1 + 435d

    divide by 15

    2u1 + 29d = 84----------------------eqn(2)

    subtract (1) from (2)

    10d = 30

    d = 3

    substitute d value in eqn (1)

    2u1 + 57 = 54

    2u1 = -3

    u1 = -1.5

    first term = -1.5

    common difference = 3

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