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    Arithmetic progression

    The 5th term of an AP is -3. The 6th term is equal to the 1st term multiplied by (-1).
    Find the first term and the common difference.
    Find the sum of the first 25 terms of the AP.

    can anyone help me get started on this?

    many thanks
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    Quote Originally Posted by decoy808 View Post
    The 5th term of an AP is -3. The 6th term is equal to the 1st term multiplied by (-1).
    Find the first term and the common difference.
    Find the sum of the first 25 terms of the AP.

    can anyone help me get started on this?

    many thanks
    Formulae:

    nth term: U_n = a+(n-1)d

    Sum of n terms: S_n = \frac{n}{2}[2a+(n-1)d]


    We know:

    U_5 = a + 4d = -3

    U_6 = a + 5d = -a

    Solve the simultaneous equations for a and d then sub into the above
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  3. #3
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    Quote Originally Posted by decoy808 View Post
    The 5th term of an AP is -3. The 6th term is equal to the 1st term multiplied by (-1).
    Find the first term and the common difference.
    Find the sum of the first 25 terms of the AP.

    can anyone help me get started on this?

    many thanks
    Let a= 1st term
    Let d= difference common between terms.

    The sequence is..

    T_1,\ \ T_2,\ \ \ \ \ T_3,\ \ \ \ \ T_4,\ \ \ \ \ \ T_5,\ \ \ \ \ \ \  T_6,....
    a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d),....

    The 5th term (a+4d) is -3
    The 6th term (a+5d) is -a

    Then the difference between these is (a+5d)-(a+4d)=5d-4d=d=-a-(-3)=3-a
    Also, a+5d = -a, so 2a+5d = 0, so 2a = -5d = -5(3-a) = 5a-15.

    2a=5a-15
    15=5a-2a=3a

    Solving for "a" allows "d" to be determined.

    The sum of an arithmetic sequence is (number\ of\ terms)(average\ value)

    The average is \frac{1st\ term+last\ term}{2}

    Hence T_{25}=a+24d

    The first term is "a".

    \frac{(T_{25}+a)25}{2}=S_{25}
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