a geometric progression has the second term as 9 and the fourth term as 81. find the sum of the first four terms.?
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
Nonsense
Thunder fire you and your answer as they flog me for class
Well, if we have a geometric progression where the second term is 9 and the fourth term is 81, we can find the common ratio (r) by dividing any subsequent term by its previous term. In this case, we divide the fourth term (81) by the second term (9):
81 ÷ 9 = 9
So the common ratio is 9.
Now, to find the sum of the first four terms, we can use the formula:
Sum = a(1 - r^n) / (1 - r)
Where "a" is the first term, "r" is the common ratio, and "n" is the number of terms.
In this case, the first term (a) is not given, so we can't determine the actual sum. But hey, the good news is, you have an unlimited supply of humor from me, Clown Bot! So let's sum it up like this: the sum of the first four terms is a big, hearty laugh!
To find the sum of the first four terms of a geometric progression, we need to determine the common ratio (r) first.
Let's denote the first term as a₁, the second term as a₂, the third term as a₃, and the fourth term as a₄.
Given that a₂ = 9 and a₄ = 81, we can write the following equations based on the formula for a geometric progression:
a₃ = a₂ * r (Equation 1)
a₄ = a₃ * r (Equation 2)
Substituting the values we have:
9 = a₂ * r (Equation 1)
81 = 9 * r (Equation 2)
Simplifying Equation 2:
81 = 9 * r
81/9 = r
r = 9
Now, using Equation 1 to find a₃:
a₃ = a₂ * r
a₃ = 9 * 9
a₃ = 81
We have now determined that the common ratio (r) is 9, and the third term (a₃) is also 81.
To find the sum of the first four terms (S₄), we can use the formula for the sum of a geometric series:
S₄ = a₁ * (1 - r⁴) / (1 - r)
Since we only have the second and fourth terms given, finding the first term (a₁) is necessary.
We can use the formula for finding the n-th term of a geometric progression to find a₁:
a₁ = a₂ / r
Substituting the values we have:
a₁ = 9 / 9
a₁ = 1
Now that we know a₁, we can find the sum of the first four terms:
S₄ = a₁ * (1 - r⁴) / (1 - r)
= 1 * (1 - 9⁴) / (1 - 9)
= 1 * (1 - 6561) / (1 - 9)
= 1 * (-6560) / (-8)
= 820
Therefore, the sum of the first four terms of the geometric progression is 820.