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    Can someone help me answer part c of this question? Question is in attachment!

    Thanks in advance!
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    (Original post by mujahid_e3)
    Can someone help me answer part c of this question? Question is in attachment!

    Thanks in advance!

    Use the formula n/2{2A + (n-1)d} = Sum (twice Jill's value)

    n = number of terms
    A= 1st term
    d = common difference

    sub in the values you know you get:

    20/2{2A + 19 x 30} = 9800
    10{2A +570} = 9800
    2A +570 = 980
    2A = 410
    A = 205
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    Use the formula S_n=\frac{1}{2}n\{2a+(n-1)d\}.

    In this case, S_n is double your answer to part (b), a is A, n is 20 and d is 30.

    Then you can rearrange to find A.
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    (Original post by ttoby)
    Use the formula S_n=\frac{1}{2}n\{2a+(n-1)d\}.

    In this case, S_n is double your answer to part (b), a is A, n is 20 and d is 30.

    Then you can rearrange to find A.
    i am finding it hard to rearrange and find a so if you can please show step by step!

    Thanks
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    (Original post by fudgesundae)
    Use the formula

    n = number of terms
    A= 1st term
    d = common difference

    sub in the values you know you get:

    20/2{2A + 19 x 30} = 9800
    10{2A +570} = 9800
    2A +570 = 980
    2A = 410
    A = 205
    ^^^^^^^
 
 
 
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