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
