Question

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

Answer 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

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

so
( - 3 )⁴ = 3⁴
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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

Answer 2

Nonsense

Answer 3

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