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Arithmetic Progression

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In an AP the 6th term is half the 4th term and the 3rd term is 15.

a. Find the fist term and the common difference.

b. How many times are needed to give a sum that is less than 65?

  • Arithmetic Progression - ,

    Arithmetic progression :

    an = a1 + ( n - 1 ) * d

    In this case :

    a6 = a1 + ( 6 - 1 ) * d

    a6 = a1 + 5 d


    a4 = a1 + ( 4 - 1 ) *d

    a4 = a1 + 3 d


    a6 = ( 1 / 2 ) a4

    a6 = ( 1 / 2 ) ( a1 + 3 d )

    a6 = a1 / 2 + 3 d / 2


    a6 = a6


    a1 + 5 d = a1 / 2 + 3 d / 2 Substract 5 d to both sides

    a1 + 5 d - 5 d = a1 / 2 + 3 d / 2 - 5 d

    a1 = a1 / 2 + 3 d / 2 - 10 d / 2

    a1 = a1 / 2 - 7 d / 2 Substract a1 / 2 to both sides

    a1 - a1 / 2 = a1 / 2 - 7 d / 2 - a1 / 2

    a1 / 2 = - 7 d / 2 Multiply both sides by 2

    a1 = - 7 d


    a3 = a1 + ( 3 - 1 ) * d

    a3 = a1 + 2 d


    a3 = 15

    15 = a1 + 2 d

    15 = - 7 d + 2 d

    15 = - 5 d Divide both sides by - 5

    15 / - 5 = d

    - 3 = d

    d = - 3


    a1 = - 7 d

    a1 = - 7 * ( - 3 ) = 21



    a1 = 21

    a2 = a1 + ( 2 - 1 ) * d = a1 + d = 21 + ( - 3 ) = 21 - 3 = 18

    a3 = a1 + ( 3 - 1 ) * d = a1 + 2 d = 21 + 2 * ( - 3 ) = 21 + ( - 6 ) = 21 - 6 = 15

    a4 = a1 + ( 4 - 1 ) * d = a1 + 3 d = 21 + 3 * ( - 3 ) = 21 + ( - 9 ) = 21 - 9 = 12

    a5 = a1 + ( 5 - 1 ) * d = a1 + 4 d = 21 + 4 * ( - 3 ) = 21 + ( - 12 ) = 21 - 12 = 9

    a6 = a1 + ( 6 - 1 ) * d = a1 + 5 d = 21 + 5 * ( - 3 ) = 21 + ( - 15 ) = 21 - 15 = 6


    AP:

    21 , 18 , 15 , 12 , 9 , 6



    b.

    The sum of the n members of a arithmetic progression :

    Sn = ( n / 2 ) ( a1 + an )

    In this case :

    65 < ( n / 2 ) ( a1 + an )

    65 < ( n / 2 ) * ( 21 + an ) Multiply both sides by 2

    2 * 65 < 2 * ( n / 2 ) * ( 21 + an )

    135 < n * ( 21 + an )


    Sn = ( n / 2 ) ( a1 + an )


    For n = 6

    S6 = ( 6 / 2 ) ( 21 + 6 )

    S6 = 3 * 27 = 81

    81 > 65 That is not solution


    For n = 5

    S5 = ( 5 / 2 ) ( 21 + 9 )

    S5 = ( 5 / 2 ) * 30 = 150 / 2 = 75

    75 > 65 That is not solution


    For n = 4

    S4 = ( 4 / 2 ) ( 21 + 12 )

    S4 = 2 * 33 = 66

    66 > 65 That is not solution.


    For n = 3

    S3 = ( 3 / 2 ) ( 21 + 15 )

    S3 = ( 3 / 2 ) * 36 = 108 / 2 = 54

    54 < 65 That is solution.


    a1 + a2 + a3 = 21 + 18 + 15 = 54

  • Arithmetic Progression - ,

    "the 6th term is half the 4th term" ---> a+5d = (1/2)(a+3d)
    2a + 10d = a + 3d
    a + 7d = 0

    "the 3rd term is 15. " ---> a + 2d = 15
    subtract those two equations ...
    5d = -15
    d = -3
    sub into
    a+ 2d = 15 ---> a - 6 = 15
    a = 21
    a = 21 , b = -3

    let the number of terms be n
    (n/2)[42 - (n-1)(-3) < 65
    n [ 42 + 3n - 3 ] < 130
    3n^2 + 39n -130 < 0

    Consider 3n^2 + 39n - 130 = 0
    n = 2.75 or a negative
    but n must be a whole number

    let n=3
    21 + 18 + 15 < 65

    3 terms are needed

  • Arithmetic Progression - ,

    nth term of 9,12,15,18

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