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HelpArithmetic progression

  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


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    You cannot find numerical values for the first two terms until you find all the terms needed to satisfy the sum. Write the difference between the first two terms as a function of x, and use that to write the difference between the first and last term. Then use what you know about how to sum an arithmetic progression.
  4. Apr 30, 2005 #3
    is the difference -2x?
  5. Apr 30, 2005 #4
    even with the common factor the sum doesn't = 310 .

    Is the question incorrect?
  6. Apr 30, 2005 #5


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    Sorry. I may have misread the expressions.

    [tex] (1 - x)^2 = 1 - 2x + x^2 [/tex]

    [tex] 1 + x^2 - (1 - x)^2 = 1 + x^2 - 1 + 2x - x^2 = 2x [/tex]

    Is this what you have?
    Last edited: Apr 30, 2005
  7. Apr 30, 2005 #6
    Yes, but i got -2x

    if this is the difference can i substitute it into

    (1-x)^2 + ( 9 x - 2x) = 10 the term

    then solve for x before sub into

    S10 = 10 x (A + L) / 2

  8. Apr 30, 2005 #7


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    the notation is confusing. I think you mean

    (1-x)^2 + (9)(- 2x) = the 10th term

    but the difference is the second term minus the first term. Get the sign of the difference right. Then substitute that expression for the 10th term into your expression for the sum and set the sum = 310. Then solve for x.
  9. Apr 30, 2005 #8
    Thank you...
  10. Apr 30, 2005 #9
    i was going down the right track...but must have lost myself with the wrong sign.....
    thanks :)
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