COMEDK 2024 Shift 1 Paper

Let's start with the given conditions:
1. b is the arithmetic mean of a and c :
b=‌

a+c

2

2. (b−a),(c−b),a are in geometric progression. In a geometric progression, the ratio of successive terms is constant. Let this common ratio be r.
Therefore, we can write:
‌

c−b

b−a

=‌

a

c−b

Cross multiplying gives us:
(c−b)2=(b−a)⋅a
Now let's substitute b=‌

a+c

2

into the geometric progression condition.
Calculating b−a and c−b :
‌b−a=‌

a+c

2

−a=‌

c−a

2

‌c−b=c−‌

a+c

2

=‌

c−a

2

Since b−a and c−b are equal, let's substitute these into the geometric progression condition:
(‌

c−a

2

)2=(‌

c−a

2

)⋅a

Simplifying, we get:
‌

(c−a)2

4

=‌

a(c−a)

2

Multiplying both sides by 4 to clear the denominators:
(c−a)2=2a(c−a)
Expanding and simplifying the equation:
c2−2ac+a2=2ac−2a2
Combining like terms, we get:
c2−4ac+3a2=0

This is a quadratic equation in terms of c. Let's solve for c :
‌c=‌

4a±√(4a)2−4⋅3a2

2

‌c=‌

4a±√16a2−12a2

2

‌c=‌

4a±√4a2

2

‌c=‌

4a±2a

2

This gives us two possible solutions for c :
‌c=‌

4a+2a

2

=3a
‌c=‌

4a−2a

2

=a

Since a,b, and c are three unequal numbers, we discard the solution c=a and accept c=3a.
Now we have the relationships a,b,c such that:
‌b=‌

a+3a

2

=2a
‌c=3a

Thus, the ratio a:b:c is:
a:2a:3a
Simplifying, we get the ratio:
1:2:3
Therefore, the correct answer is Option C:1:2:3.