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Math Help - [SOLVED] Arithmetic and Geometric Progressions Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Arithmetic and Geometric Progressions Question

    Question:
    The 1st term of an arithmetic progression is a and the common difference is d, where d \neq 0.

    (i) Write down expressions, in terms of a and d, for the 5th term and 15th term.

    The 1st term, the 5th term and the 15th term of the arithmetic progression are the first terms of a geometric progression.

    (ii) Show that 3a = 8d.
    (iii) Find the common ratio of the geometric progression.


    Attempt:

    (i) U_5 = a + 4d
    U_{15} = a + 14d

    (ii) Arithmetic Progression
    U_5 = a + 4d
    U_{15} = a + 14d

    Geometric Progression
    U_2 = a.r
    U_3 = a.r^2

    I don't know what to do next... Need help!
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  2. #2
    Moo
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    Hello,

    What you did was good

    Let's rename the geometric series into S :

    S_2 = a.r
    S_3=a.r^2


    Now the text says that S_2=U_5 and S_3=U_{15}

    Hence you have to solve this :

    \begin{bmatrix} a+4d=ar \\ a+14d=ar^2 \end{bmatrix}

    Can you try to get 3a=8d ?
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  3. #3
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    Quote Originally Posted by looi76 View Post
    Question:
    The 1st term of an arithmetic progression is a and the common difference is d, where d \neq 0.

    (i) Write down expressions, in terms of a and d, for the 5th term and 15th term.

    The 1st term, the 5th term and the 15th term of the arithmetic progression are the first terms of a geometric progression.

    (ii) Show that 3a = 8d.
    (iii) Find the common ratio of the geometric progression.

    Attempt:

    (i) U_5 = a + 4d
    U_{15} = a + 14d

    (ii) Arithmetic Progression
    U_5 = a + 4d
    U_{15} = a + 14d

    Geometric Progression
    U_2 = a.r
    U_3 = a.r^2

    I don't know what to do next... Need help!
    a + 4d = ar
    a + 14d = ar^2

    ar + 4dr = ar^2
    a + 4d + 4dr = ar^2
    a + 4d + 4dr = a + 14d
    4d + 4dr = 14d
    4dr = 10d
    r = \frac{10d}{4d} = \frac{5}{2}

    a + 4d = \frac{5}{2}a
    4d = \frac{3}{2}a
    8d = 3a
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  4. #4
    Member looi76's Avatar
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    Quote Originally Posted by icemanfan View Post
    a + 4d = ar
    a + 14d = ar^2

    ar + 4dr = ar^2
    \color{red}{a + 4d + 4dr = ar^2}
    a + 4d + 4dr = a + 14d
    4d + 4dr = 14d
    4dr = 10d
    r = \frac{10d}{4d} = \frac{5}{2}

    a + 4d = \frac{5}{2}a
    4d = \frac{3}{2}a
    8d = 3a
    How did you get the equation in red?
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  5. #5
    Moo
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    ar=a+4d
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  6. #6
    Member looi76's Avatar
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    Quote Originally Posted by Moo View Post
    ar=a+4d
    Tell now I didn't get it. How did u get that equation?!
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  7. #7
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    Quote Originally Posted by looi76 View Post
    Tell now I didn't get it. How did u get that equation?!

    \color{blue}{a + 4d = ar}
    a + 14d = ar^2

    {\color{red}{ar}} + 4dr = ar^2
    But from the blue equation above this ar = a+4d, so replace the red
    \color{red}ar with \color{blue}a+4d

    {\color{blue}{a + 4d}} + 4dr = ar^2

    Got it ?
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