Geometric sequence: 2b+2,b+4,b,

by Sylvia Andrew
(Kota kinabalu)

Given the first three terms of Geometric sequence: 2b+2,b+4,b, where each of the term is positive, find:
i) the value of b
ii) the first term and common ratio

Comments for Geometric sequence: 2b+2,b+4,b,

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Jul 21, 2011

Geometric Sequence
by: Staff

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Part II

solve this equation for b

b = (2b+2) * [(b+4)/ (2b+2)]^(2)

b = (2b+2) * (b+4) * (b+4) / [(2b+2) *(2b+2)]

b = [(2b+2)/(2b+2)] * (b+4) * (b+4) / (2b+2)

b = 1 * (b+4) * (b+4) / (2b+2)

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

b * (2b+2) = [(b+4) * (b+4) / (2b+2)] * (2b+2)

b * (2b+2) = (b+4) * (b+4) * [(2b+2) / (2b+2)]

b * (2b+2) = (b+4) * (b+4) * 1

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

2b² + 2b = (b+4) * (b+4)

2b² + 2b = b² + 8b + 16

2b² - b² + 2b = b² - b² + 8b + 16

b² + 2b = b² - b² + 8b + 16

b² + 2b = 0 + 8b + 16

b² + 2b = 8b + 16

b² + 2b - 8b = 8b - 8b + 16

b² - 6b = 8b - 8b + 16

b² - 6b = 0 + 16

b² - 6b = 16

b² - 6b - 16 = 16 - 16

b² - 6b - 16 = 0

(b-8)*(b+2) = 0

(b-8)*(b+2) / (b+2) = 0 /(b+2)

(b-8)*1= 0

(b-8) = 0

b - 8 = 0

b - 8 + 8 = 0 + 8

b + 0 = 0 + 8

b = 8

(b-8)*(b+2) / (b-8) = 0 /(b-8)

(b-8)*(b+2) / (b-8) = 0

[(b-8)/(b-8)]*(b+2) = 0

1*(b+2) = 0

(b+2) = 0

b + 2 = 0

b + 2 - 2 = 0 - 2

b + 0 = 0 - 2

b = 0 - 2

b = -2

the value of b ∈{-2, 8}

However, since each term must be positive, b must = 8

Calculate the first term

If n = 1, the first term is 2b+2

x_1 = 2b + 2

x_1 = 2*8 + 2

x_1 = 16 + 2

x_1 = 18

The common ratio
r = (b+4)/ (2b+2)
r = (8+4)/ (2*8+2)
r = (12)/ (18)
r = (6)/ (9)
r = 2/ 3

the final answer:

i) the value of b = 8

ii) the first term, x_1 = 18

iii) the common ratio, r = 2/ 3

verify b = 8 by substituting 8 for the value of b in the equation

b = (2b+2) * [(b+4)/ (2b+2)]^(2)

8 = (2*8 + 2) * [(8+4)/ (2*8 + 2)]^(2)

8 = (16 + 2) * [(12)/ (16 + 2)]^(2)

8 = (18) * [(12)/ (18)]^(2)

8 = (18) * (2/3)^(2)

8 = (18) * (4/9)

8 = (18/9) * (4)

8 = (2) * (4)

8 = 8, OK

Thanks for writing.

Staff
www.solving-math-problems.com

Jul 21, 2011

Geometric Sequence
by: Staff

Part I

The question:

by Sylvia Andrew
(Kota kinabalu)

Given the first three terms of Geometric sequence: 2b+2,b+4,b, where each of the term is positive, find:

i) the value of b
ii) the first term and common ratio

The answer:

A geometric sequence has the (general) form:

x_n = x_1 * (r)^(n - 1)

x_n = x with a subscript of n (this is the nth term in the sequence)

x_1 = x with a subscript of 1 (this is the 1st term in the sequence)

n = number of terms

r = the common ratio

r, the common ratio, can be calculated as follows:
r_n = x_n/x_n-1

(n must be greater than 1)

r_n = r with a subscript of n (this is the common ratio)

x_n = x with a subscript of n (this is the nth term in the sequence)

x_n-1 = x with a subscript of n-1 (this is the n-1 term in the sequence)

r_2 = x_2/x_1

r_2 = (b+4)/ (2b+2)

The common ratio, r = (b+4)/ (2b+2)

The first three terms in the geometric sequence:

x_n = x_1 * (r)^(n - 1)

If n = 1, the first term is 2b+2

x_n = (2b+2) * [(b+4)/ (2b+2)]^(n - 1)

x_1 = (2b+2) * [(b+4)/ (2b+2)]^(1 - 1)

x_1 = (2b+2) * [(b+4)/ (2b+2)]^(0)

x_1 = (2b+2) * 1

2b+2 = 2b+2

If n = 2, the second term is b+4

x_n = (2b+2) * [(b+4)/ (2b+2)]^(n - 1)

x_2 = (2b+2) * [(b+4)/ (2b+2)]^(2 - 1)

x_2 = (2b+2) * [(b+4)/ (2b+2)]^(1)

x_2 = (2b+2) * [(b+4)/ (2b+2)]

b+4 = (2b+2) * [(b+4)/ (2b+2)]

b+4 = b+4

If n = 3, the third term is b

x_n = (2b+2) * [(b+4)/ (2b+2)]^(n - 1)

x_3 = (2b+2) * [(b+4)/ (2b+2)]^(3 - 1)

x_3 = (2b+2) * [(b+4)/ (2b+2)]^(2)

b = (2b+2) * [(b+4)/ (2b+2)]^(2)
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Geometric sequence: 2b+2,b+4,b,

Comments for Geometric sequence: 2b+2,b+4,b,