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a,b,c in AP and b-a,c-b,a in GP: Find a:b:c | JEE

JEE Maths question with a full step-by-step solution.

Question
If a,b,ca, b, c are three distinct numbers such that a,b,ca, b, c are in A.P. and b−a, c−b, ab-a,\ c-b,\ a are in G.P., then a:b:ca:b:c is
A2:3:42:3:4
B3:4:53:4:5
C1:3:51:3:5
D1:2:31:2:3correct
Solution
Step 1: A.P. gives 2b=a+c2b=a+c and equal differences b−a=c−b.b-a=c-b. Step 2: G.P. of b−a, c−b, ab-a,\ c-b,\ a gives (c−b)2=a(b−a)(c-b)^2=a(b-a). Replace c−bc-b by b−ab-a:
(b−a)2=a(b−a) ⇒ b−a=a.(b-a)^2=a(b-a)\ \Rightarrow\ b-a=a.
 Step 3: So b=2a, c=3a⇒a:b:c=1:2:3.b=2a,\ c=3a\Rightarrow a:b:c=1:2:3. Correct answer: (4)
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