need help on math question?

determine the sum of the arithmetic series

9 + 21 + 33 +... + 693

thanks in advance..

4 Answers

Relevance
  • Anonymous
    8 years ago
    Favorite Answer

    In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant

    Sum=(a1+an)*n/2, where a1-first member of an arithmetic progression, an - the last member of an arithmetic progression, n - number of members of an arithmetic progression

    an=a1+(n-1)d, where d - the difference between the members of an arithmetic progression

    In your task a1=9, d=a2-a1=21-9=12, an=693

    Therefore n=(an-a1)/d+1=(693-9)/12+1=58

    Then Sum=(9+693)*58/2=20358

    Answer: Sum=20358

    • Commenter avatarLogin to reply the answers
  • 8 years ago

    common difference, d= 12.

    last term, an= a + (n-1)*d

    693 = 9 + (n-1)*12 => n= 58.

    sum, S= (n/2)* [ (a+an)/2 ]

    => S= (58/2) * [ (9 + 693)/ 2 ]

    calculate :-)

    • Commenter avatarLogin to reply the answers
  • 8 years ago

    a = 9 and d = 12

    => nth term is a + (n - 1)d = 9 + 12(n - 1) = 12n - 3

    693 = 12n - 3

    => 12n = 696

    i.e. n = 696/12 = 58.....i.e. 58 terms

    => Sn = (n/2)[2a + (n - 1)d]

    i.e. S58 = (58/2)[18 + (57)(12)] = 20,358

    :)>

    • Commenter avatarLogin to reply the answers
  • 8 years ago

    add each number with 12 and then the answer again with 12..

    but this will be right only if the last number is 633..

    i thing you have made mistake in numbers

    9+21+33+45+57+69+.........+633

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.