a. The common ratio is 2.
Given that the sum of the first five terms of the geometric series is 33, we use the formula for the sum of terms of a geometric series, S
Sₙ = a(rⁿ - 1)/(r - 1) where a = first term and r = common ratio
Since n = 5, the first 5 terms, and S₅ = 33
Sₙ = a(rⁿ - 1)/(r - 1)
33 = a(r⁵ - 1)/(r - 1) (1)
Also, when n = 10, the sum of the first 10 terms is S₁₀ = -1023
So, -1023 = a(r¹⁰ - 1)/(r - 1) (2)
Dividing (2) by (1), we have
-1023/33 = a(r¹⁰ - 1)/(r - 1) ÷ a(r⁵ - 1)/(r - 1)
-31 = (r¹⁰ - 1)/(r⁵ - 1)
-31(r⁵ - 1) = r¹⁰ - 1
-31r⁵ + 31 = r¹⁰ - 1
r¹⁰ - 1 + 31r⁵ - 31 = 0
r¹⁰ + 31r⁵ - 32 = 0
Let r⁵ = y
(r⁵)² + 31r⁵ - 32 = 0
y² + 31y - 32 = 0
Factorizing, we have
y² + 32y - y - 32 = 0
y(y + 32) - (y + 32) = 0
(y - 1)(y - 32) = 0
y - 1 = 0 or y - 32 = 0
y = 1 or y = 32
r⁵ = 1 or r⁵ = 32
r = ⁵√1 or r = ⁵√32
r = 1 or r = 2
Since for a geometric series, r ≠ 1, r = 2.
So, the common ratio is 2.
ii. The first term of the series.
The first term of the series is 33/31
Using (1)
33 = a(r⁵ - 1)/(r - 1) (1) where r = 2,
33 = a(2⁵ - 1)/(2 - 1) (1)
33 = a(32 - 1)/1
33 = 31a
a = 33/31
So, the first term of the series is 33/31
b. Find the general term of the series. Simplify your answer.
The general term of the geometric series is (33/62) × 2ⁿ
The general term of a geometric series is Uₙ = arⁿ⁻¹
With a = 33/31 and r = 2,
Uₙ = arⁿ⁻¹
Uₙ = (33/31) × 2ⁿ⁻¹
U��� = (33/31) × 2ⁿ/2
Uₙ = (33/62) × 2ⁿ
So, the general term of the geometric series is (33/62) × 2ⁿ
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