Question

If $a,b,c$ are in $AP,b-a,c-b$ and $a$ are in GP, then $a:b:c$ is

• A. 1 : 2 : 3
• B. 1 : 3 : 5
• C. 2 : 3 : 4
• D. 1 : 2 : 4

Answer 1

Answer: (A)

Given, that, $2b=a+c$
and $(c-b{)}^2=(b-a)a$
$\therefore (b-a{)}^2=(b-a)a$
$\Rightarrow b=2a$
$\Rightarrow c=3a$
$\Rightarrow a:b:c=1:2:3$