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a geometric progression has the second term as 9 and the fourth term as 81. find the sum of the first four terms.?

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3 answers
  1. a1 = first term

    r = common ratio

    Sn = nth partial sum of a geometric sequence

    an = a1 ∙ r ⁿ⁻¹

    a2 = a1 ∙ r ²⁻¹ = a1 ∙ r¹ = a1 ∙ r = 9

    a4 = a1 ∙ r ⁴⁻¹ = a1 ∙ r³ = 81

    a4 / a2 = a1 ∙ r³ / a1 ∙ r = r²

    r² = 81 / 9 = 9

    r = √ r²

    r = ± √ 9

    r = ± 3

    If:

    a2 = a1 ∙ r = 9

    then

    a1 = 9 / r

    a1 = 9 / ± 3

    a1 = ± 3

    Sn = a1 ∙ ( 1 - rⁿ ) / ( 1 - r )

    S4 = a1 ∙ ( 1 - r⁴ ) / ( 1 - r )

    There are 4 possible cases:

    1.

    a1 = 3 , r = 3

    2.

    a1 = 3 , r = - 3

    3.

    a1 = - 3 , r = 3

    4.

    a1 = - 3 , r = - 3

    __________________________________________
    Remark:
    ( - 3 )⁴ = ( - 3 )⁴ = [ ( - 1 ) ∙ 3 ]⁴ = ( - 1 )⁴ ∙ 3⁴ = 1 ∙ 3⁴ =3⁴

    so
    ( - 3 )⁴ = 3⁴
    ___________________________________________

    First case:

    a1 = 3 , r = 3

    S4 = a1 ∙ ( 1 - r⁴ ) / ( 1 - r )

    S4 = 3 ∙ ( 1 - 3⁴ ) / ( 1 - 3 )

    S4 = 3 ∙ ( 1 - 81 ) / ( - 2 )

    S4 = 3 ∙ ( - 80 ) / ( - 2 )

    S4 = 3 ∙ 40

    S4 = 120

    Second case:

    a1 = 3 , r = - 3

    S4 = a1 ∙ ( 1 - r⁴ ) / ( 1 - r )

    S4 = 3 ∙ [ 1 - ( - 3)⁴ ] / [ 1 - ( - 3 ) ]

    S4 = 3 ∙ ( 1 -- 3⁴ ) / [ 1 - ( - 3 ) ]

    S4 = 3 ∙ ( 1 - 81 ) / ( 1 + 3 )

    S4 = 3 ∙ ( - 80 ) / 4

    S4 = 3 ∙ ( - 20 )

    S4 = - 60

    Third case:

    a1 = - 3 , r = 3

    S4 = a1 ∙ ( 1 - r⁴ ) / ( 1 - r )

    S4 = - 3 ∙ ( 1 - 3⁴ ) / ( 1 - 3 )

    S4 = - 3 ∙ ( 1 - 81 ) / ( - 2 )

    S4 = - 3 ∙ ( - 80 ) / ( - 2 )

    S4 = - 3 ∙ 40

    S4 = - 120

    Fourth case:

    a1 = - 3 , r = - 3

    S4 = a1 ∙ ( 1 - r⁴ ) / ( 1 - r )

    S4 = - 3 ∙ [ 1 - ( - 3 )⁴ ] / [ 1 - ( - 3 ) ]

    S4 = - 3 ∙ ( 1 - 3⁴ ) / ( 1 + 3 )

    S4 = - 3 ∙ ( 1 - 81 ) / 4

    S4 = - 3 ∙ ( - 80 ) / 4

    S4 = - 3 ∙ ( - 20 )

    S4 = 60

    So:

    S4 = ± 60

    OR

    S4 = ± 120

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  2. Nonsense

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  3. Thunder fire you and your answer as they flog me for class

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