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Arithmetic progression Trouble

  1. Apr 30, 2005 #1
    The first term of an arithmetic progression is (1-x)^2 and the second term is 1+x^2 .If the sumj of the first ten terms is 310 , find the possible values of x.

    I have my A/S maths exam next month, but i am still having trouble with arithmetic progression. The above question is causing me some trouble .

    First i expanded the brackets using binomial expansion .

    Then as i had a quadratic i used the theorem to find values for x .

    Once i found x i substituted into the first two terms to find the difference .

    I found the first term = 1.98 the second = 6.8 with a diff of 4.8 .

    As a + ( 9 X d ) = the tenth term = 45.36

    And the formula for the sum is

    S 10 = 10 x ( a + l)/2 .....where l = 45.36

    Why do i keep getting 236 .7

    Am i doing something drasticaly wrong?

    Many thanks .
     
  2. jcsd
  3. Apr 30, 2005 #2

    arildno

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