COMEDK2024Morning ShiftMathematicsSequences and SeriesActual
Given a, b, c are three unequal numbers such that b is arithmetic mean of a and c and (b-a),(c-b), a are in geometric progression. Then a: b: c is
Options
- A1: 3: 5
- B2: 3: 5
- C1: 2: 3
- D1: 2: 4
Correct answer
C. 1: 2: 3
Step-by-step solution
Given that b is the arithmetic mean of a and c , we have b = a+c 2 , which implies 2b = a+c or c = 2b - a . The terms (b-a), (c-b), a are in geometric progression. Therefore, (c-b)^2 = (b-a) a . Substituting c = 2b - a into the geometric progression condition: (2b - a - b)^2 = (b-a)a (b-a)^2 = (b-a)a Since a, b, c are unequal, b a , so we can divide by (b-a) : b-a = a b = 2a Now substitute b = 2a into the relation c = 2b - a : c = 2(2a) - a = 4a - a = 3a The ratio a : b : c is a : 2a : 3a , which simplifies to 1 :