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The sum of integrs from 40-60, inclusive, is 1050. What is the aum of the integers from 60-80, inclusive?

I figured out the answer by adding them with a calculator but i was wondering if there was an algebraic way of solving this?

  • Math -

    Since you are adding 20 to each number, just add 20*21 to the total.

    Or, thinking of the values as an arithmetic progression, recall that the sum of n terms is

    Sn = n/2 (T1+Tn)
    With a=40,d=1,

    S21 = 21/2 (40+60) = 1050

    With a=60,d=1,

    S21 = 21/2 (60+80) = 1470

    Note that 1470 = 1050 + 21*20

  • Math -

    Ok, thank you!

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