# 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,

 Jul 21, 2011 Geometric Sequence by: Staff --------------------------------------------------------- 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) ---------------------------------------------------------