neildvesterdidi
Answered

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Given the first three terms of Geometric Sequence: 2b+2, b+4, b, where each of the tem is
positive, find
1) the value of b. (8)
ii) the first term and common ratio. (18,2/3)
(2009)​

Sagot :

Eduard22sly

Answer:

1. b = 8

2. First term (T₁) = 18

Common ratio (r) = 2/3

Step-by-step explanation:

1. Determination of the value of b.

2b+2, b+4, b

First term (T₁) = 2b + 2

2nd term (T₂) = b + 4

3rd term (T₃) = b

The value of b can be obtained as follow:

Common ratio = T₂/T₁ = T₃/T₂

T₂/T₁ = T₃/T₂

(b + 4)/(2b + 2) = b/(b + 4)

Cross multiply

(b + 4)(b + 4) = b(2b + 2)

Expand

b(b + 4) + 4(b + 4) = b(2b + 2)

b² + 4b + 4b + 16 = 2b² + 2b

b² + 8b + 16 = 2b² + 2b

Rearrange

2b² – b² + 2b – 8b – 16 = 0

b² – 6b – 16 = 0

Solving by factorisation

b² – 8b + 2b – 16 = 0

b(b – 8) + 2(b – 8) = 0

(b – 8)(b + 2) = 0

b – 8 = 0 or b + 2 = 0

b = 8 or b = –2

Since each of the terms are positive, therefore, b is 8.

2. Determination of the first term and common ratio.

2b+2, b+4, b

First term (T₁) = 2b + 2

b = 8

First term (T₁) = 2(8) + 2

First term (T₁) = 16 + 2

First term (T₁) = 18

2nd term (T₂) = b + 4

b = 8

2nd term (T₂) = 8 + 4

2nd term (T₂) = 12

Common ratio (r) = T₂/T₁

First term (T₁) = 18

2nd term (T₂) = 12

Common ratio (r) = 12/18

Common ratio (r) = 2/3

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