# Three Numbers in Geometric Progression - Maths

by Mohan Kumar
(Rajaji Nagar)

three numbers in geometric progression, their sum is 7 and the sum their reciprocals is 7/5 find the terms. Search instead for three numbers in geometric progression, thier sum is 7 and the sum thier reciprocals is 7/5 find the terms

### Comments for Three Numbers in Geometric Progression - Maths

 Apr 05, 2012 Three Numbers in Geometric Progression by: Staff Part I Question: by Mohan Kumar (Rajaji Nagar) three numbers in geometric progression, their sum is 7 and the sum their reciprocals is 7/5 find the terms. Search instead for three numbers in geometric progression, their sum is 7 and the sum their reciprocals is 7/5 find the terms Answer: A geometric sequence has the (general) form: a_n = a_1 * (r)^(n - 1) a_n = a with a subscript of n (this is the nth term in the sequence) a_1 = a with a subscript of 1 (this is the 1st term in the sequence) n = number of terms r = the common ratio The three numbers are: 1st number: a_1 2nd number: a_2 3rd number: a_3 a_1 = a_1 a_2 = a_1 * (r)^(2 - 1) = a_1 * (r)^(1) = a_1 * (r) a_3 = a_1 * (r)^(3 - 1) = a_1 * (r)^(2) Since the sum of the three numbers is 7 a_1 + a_2 + a_3 = 7 a_1 + a_1 * (r) + a_1 * (r)^(2) = 7 a_1(1 + r + r^2) = 7 Since the sum of the three reciprocals is: (1/a_1) + (1/a_2) + (1/a_3) = 7/5 (1/a_1) + (1/[a_1 * (r)] + (1/ [a_1 * (r)^(2)] = 7/5 (1 + r + r²) / (a_1 * r²) = 7/5 There are two equations with two unknowns. The next step is to solve the equations for the two unknowns. Divide the 1st equation (sum of the three numbers = 7) by the 2nd equation (sum of the three reciprocals = 7/5) [a_1(1 + r + r^2) = 7] ÷ [(1 + r + r²) / (a_1 * r²) = 7/5] [a_1(1 + r + r^2)] / [(1 + r + r²) / (a_1 * r²)] = 7 / (7/5) [a_1 * (1 + r + r²) * (a_1 * r²)] / (1 + r + r²) = (7 * 5) / 7 [a_1 * (a_1 * r²) * (1 + r + r²) ] / (1 + r + r²) = (7 * 5) / 7 [a_1 * (a_1 * r²)] * [(1 + r + r²) / (1 + r + r²)] = 5 * (7 / 7) [a_1 * (a_1 * r²)] * (1) = 5 * (1) [a_1 * (a_1 * r²)] = 5 (a_1)² * r² = 5 (a_1 * r)² = 5 √(a_1 * r)² = √5 a_1 * r = √5 a_1 * r = ±√5 a_1 = (±√5)/r Substitute value for a_1 value into the 1st equation a_1(1 + r + r^2) = 7 [(±√5)/r ] * (1 + r + r^2) = 7 Solve for r (±√5) * [(1 + r + r²) /r ] = 7 (±√5) * [(1 + r + r²) /r ] = 7 (±√5) * (1 + r + r²) = 7*r ±√5 + (±√5)r + (±√5)r² = 7r ±√5 + (±√5)r + (±√5)r² - 7r = 0 (±√5)r² + (±√5)r - 7r ±√5 = 0 (±√5)r² + [±√(5) - 7]r ±√5 = 0 This is a quadratic equation ----------------------------------------------------------------------

 Apr 05, 2012 Three Numbers in Geometric Progression by: Staff ---------------------------------------------------------------------- Part II Using the quadratic formula to solve it is probably the fastest way: ax² + bx + c = 0 x = [-b ± √(b² - 4ac)]/(2a) (±√5)r² + [±√(5) - 7]r ±√5 = 0 a = (±√5) b = [±√(5) - 7] c = ±√5 to make the calculations a little easier, I am going to evaluate √5 ±√5 = ±2.2360679774998 = ±2.24 a = 2.24 b = -4.76 c = 2.24 r ∈ {0.70346, 1.4215} if r = 0.70346 a_1 = √(5)/r a_1 = 2.24/0.70346 = 3.18426 a_2 = a_1 * (r) a_2 = 3.18426 * (0.70346) = 2.2399995396 a_3 = a_1 * r² a_3 = 3.18426 * (0.70346)² = 1.575750076127 The sum of the three terms = 7.0 a_1 + a_2 + a_3 = 3.18426 + 2.2399995396+ 1.575750076127 = 7.0 the sum of the reciprocals of all three terms = 7/5 (1/a_1) + (1/a_2) + (1/a_3) = 1/3.18426 + 1/2.2399995396 + 1/1.575750076127 = 1.4 = 7/5 ------------------------------------------------------------------ if r = 1.4215 a_1 = √(5)/r a_1 = 2.24/1.4215 = 1.5758002110447 a_2 = a_1 * (r) a_2 = 1.5758002110447* (1.4215) = 2.2399995396 a_3 = a_1 * r² a_3 = 1.5758002110447 * (1.4215)² = 3.1841600000001 The sum of the three terms = 7.0 a_1 + a_2 + a_3 = 1.5758002110447 + 2.2399995396 + 3.1841600000001 = 7.0 the sum of the reciprocals of all three terms = 7/5 (1/a_1) + (1/a_2) + (1/a_3) = 1/1.5758002110447 + 1/2.2399995396 + 1/3.1841600000001 = 1.4 = 7/5 >>> The final answer is: if r = 0.70346 terms = 3.18426, 2.2399995396, 1.575750076127 if r = 1.4215 terms = 1.5758002110447, 2.2399995396, 3.1841600000001 Thanks for writing. Staff www.solving-math-problems.com